Telangana Board

10th Maths Statistics Exercise 14.4 Solutions

Exercise 14.4 Solutions – Class X Mathematics

Exercise 14.4 Solutions

From Class X Mathematics textbook by the State Council of Educational Research and Training, Telangana, Hyderabad

Question 1

The following distribution gives the daily income of 50 workers of a factory.

Daily income (in Rupees) 250-300 300-350 350-400 400-450 450-500
Number of workers 12 14 8 6 10

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.

Solution:

Step 1: Convert to less than type cumulative frequency distribution:

Daily income less than (in Rupees) 300 350 400 450 500
Cumulative frequency 12 12+14=26 26+8=34 34+6=40 40+10=50

Question 2

During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight (in kg) Less than 38 Less than 40 Less than 42 Less than 44 Less than 46 Less than 48 Less than 50 Less than 52
Number of students 0 3 5 9 14 28 32 35

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

Solution:

Step 1: The data is already in less than type cumulative frequency form.

Step 2: Drawing the ogive:

Ogive (Less than type) for weights of students

Step 3: Finding median from the graph:

Total number of students (n) = 35

Median position = n/2 = 17.5

From the graph, the x-coordinate corresponding to y=17.5 is approximately 46.5 kg.

Step 4: Verifying using the formula:

The median class is 46-48 (since 17.5 falls in the cumulative frequency of 28)

Using the formula:

\[ \text{Median} = L + \left(\frac{\frac{n}{2} – cf}{f}\right) \times h \]

Where:
L = 46 (lower limit of median class)
cf = 14 (cumulative frequency before median class)
f = 14 (frequency of median class)
h = 2 (class width)

\[ \text{Median} = 46 + \left(\frac{17.5 – 14}{14}\right) \times 2 = 46 + \left(\frac{3.5}{14}\right) \times 2 = 46 + 0.5 = 46.5 \text{ kg} \]

This matches our graphical estimate.

Question 3

The following table gives production yield per hectare of wheat of 100 farms of a village.

Production yield (Quintal/Hectare) 50-55 55-60 60-65 65-70 70-75 75-80
Number of farmers 2 8 12 24 38 16

Change the distribution to a more than type distribution, and draw its ogive.

Solution:

Step 1: Convert to more than type cumulative frequency distribution:

Production yield more than (Quintal/Hectare) 50 55 60 65 70 75
Cumulative frequency 100 100-2=98 98-8=90 90-12=78 78-24=54 54-38=16

Step 2: Drawing the ogive (more than type):

Ogive (More than type) for production yield

10th Maths Statistics Exercise 14.3 Solutions

Exercise 14.3 Solutions – Class X Mathematics

Exercise 14.3 Solutions

Class X Mathematics – State Council of Educational Research and Training, Telangana, Hyderabad

1. Electricity Consumption Problem

Monthly consumption 65-85 85-105 105-125 125-145 145-165 165-185 185-205
Number of consumers 4 5 13 20 14 8 4

Solution:

Median:

First, we calculate cumulative frequencies:

ClassFrequencyCumulative Frequency
65-8544
85-10559
105-1251322
125-1452042
145-1651456
165-185864
185-205468

Median class is where cumulative frequency ≥ n/2 = 34 → 125-145

Using median formula:

Median = L + [(n/2 – CF)/f] × h

Where L = 125, n = 68, CF = 22, f = 20, h = 20

Median = 125 + [(34 – 22)/20] × 20 = 125 + 12 = 137

Mean:

Using assumed mean method with A = 135:

ClassMidpoint (x)Frequency (f)d = (x – A)/hf × d
65-85754-3-12
85-105955-2-10
105-12511513-1-13
125-1451352000
145-16515514114
165-1851758216
185-2051954312
Total7

Mean = A + (Σfd/Σf) × h = 135 + (7/68) × 20 ≈ 137.06

Mode:

Modal class is 125-145 (highest frequency = 20)

Using mode formula:

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where L = 125, f₁ = 20, f₀ = 13, f₂ = 14, h = 20

Mode = 125 + [(20 – 13)/(40 – 13 – 14)] × 20 ≈ 125 + (7/13) × 20 ≈ 135.77

Comparison: Mean (137.06) > Median (137) > Mode (135.77)

2. Median Problem with Missing Frequencies

Class interval 0-10 10-20 20-30 30-40 40-50 50-60
Frequency 5 x 20 15 y 5

Given median = 28.5, n = 60

Solution:

Total observations: 5 + x + 20 + 15 + y + 5 = 60 ⇒ x + y = 15

Median class is where cumulative frequency ≥ n/2 = 30 → 20-30

Using median formula:

28.5 = 20 + [(30 – (5 + x))/20] × 10

8.5 = (25 – x)/2 ⇒ 17 = 25 – x ⇒ x = 8

Since x + y = 15 ⇒ y = 7

Solution: x = 8, y = 7

3. Insurance Policy Holders Age Distribution

Age (in years) Below 20 Below 25 Below 30 Below 35 Below 40 Below 45 Below 50 Below 55 Below 60
Number of policy holders 2 6 24 45 78 89 92 98 100

Solution:

First convert to frequency distribution:

ClassFrequencyCumulative Frequency
18-2022
20-2546
25-301824
30-352145
35-403378
40-451189
45-50392
50-55698
55-602100

Median class is where cumulative frequency ≥ n/2 = 50 → 35-40

Using median formula:

Median = 35 + [(50 – 45)/33] × 5 ≈ 35 + 0.76 ≈ 35.76 years

4. Leaves Length Measurement

Length (in mm) 118-126 127-135 136-144 145-153 154-162 163-171 172-180
Number of leaves 3 5 9 12 5 4 2

Solution:

Convert to continuous classes as suggested:

ClassFrequencyCumulative Frequency
117.5-126.533
126.5-135.558
135.5-144.5917
144.5-153.51229
153.5-162.5534
162.5-171.5438
171.5-180.5240

Median class is where cumulative frequency ≥ n/2 = 20 → 144.5-153.5

Using median formula:

Median = 144.5 + [(20 – 17)/12] × 9 = 144.5 + 2.25 = 146.75 mm

5. Neon Lamps Life Time

Life time (in hours) 1500-2000 2000-2500 2500-3000 3000-3500 3500-4000 4000-4500 4500-5000
Number of lamps 14 56 60 86 74 62 48

Solution:

Calculate cumulative frequencies:

ClassFrequencyCumulative Frequency
1500-20001414
2000-25005670
2500-300060130
3000-350086216
3500-400074290
4000-450062352
4500-500048400

Median class is where cumulative frequency ≥ n/2 = 200 → 3000-3500

Using median formula:

Median = 3000 + [(200 – 130)/86] × 500 ≈ 3000 + 406.98 ≈ 3406.98 hours

6. Surnames Letters Distribution

Number of letters 1-4 4-7 7-10 10-13 13-16 16-19
Number of surnames 6 30 40 16 4 4

Solution:

Median:

First make classes continuous and calculate cumulative frequencies:

ClassFrequencyCumulative Frequency
0.5-4.566
4.5-7.53036
7.5-10.54076
10.5-13.51692
13.5-16.5496
16.5-19.54100

Median class is where cumulative frequency ≥ n/2 = 50 → 7.5-10.5

Median = 7.5 + [(50 – 36)/40] × 3 = 7.5 + 1.05 = 8.55

Mean:

Using midpoint method:

ClassMidpoint (x)Frequency (f)f × x
1-42.5615
4-75.530165
7-108.540340
10-1311.516184
13-1614.5458
16-1917.5470
Total100832

Mean = Σfx/Σf = 832/100 = 8.32

Mode:

Modal class is 7-10 (highest frequency = 40)

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where L = 7, f₁ = 40, f₀ = 30, f₂ = 16, h = 3

Mode = 7 + [(40 – 30)/(80 – 30 – 16)] × 3 ≈ 7 + (10/34) × 3 ≈ 7.88

7. Students Weight Distribution

Weight (in kg) 40-45 45-50 50-55 55-60 60-65 65-70 70-75
Number of students 2 3 8 6 6 3 2

Solution:

Calculate cumulative frequencies:

ClassFrequencyCumulative Frequency
40-4522
45-5035
50-55813
55-60619
60-65625
65-70328
70-75230

Median class is where cumulative frequency ≥ n/2 = 15 → 55-60

Using median formula:

Median = 55 + [(15 – 13)/6] × 5 ≈ 55 + 1.67 ≈ 56.67 kg

10th Maths Statistics Exercise 14.2 Solutions

Exercise 14.2 Solutions – Class X Mathematics

Exercise 14.2 Solutions

Class X Mathematics
State Council of Educational Research and Training, Telangana, Hyderabad

Problem 1

The following table shows the ages of the patients admitted in a hospital on a particular day:

Age (in years) 5-15 15-25 25-35 35-45 45-55 55-65
Number of patients 6 11 21 23 14 5

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Solution:

Mode Calculation:

The modal class is the class with the highest frequency. Here, the highest frequency is 23, which corresponds to the class 35-45.

Using the mode formula:

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where:
L = lower limit of modal class = 35
f₁ = frequency of modal class = 23
f₀ = frequency of class preceding modal class = 21
f₂ = frequency of class succeeding modal class = 14
h = class width = 10

Mode = 35 + [(23 – 21)/(2×23 – 21 – 14)] × 10
= 35 + [2/(46 – 35)] × 10
= 35 + (2/11) × 10
= 35 + 1.818 ≈ 36.82

Mean Calculation:

Class Interval Midpoint (xi) Frequency (fi) fixi
5-15 10 6 60
15-25 20 11 220
25-35 30 21 630
35-45 40 23 920
45-55 50 14 700
55-65 60 5 300
Total 80 2830

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{2830}{80} = 35.375\)

Comparison and Interpretation:

Mode = 36.82 years, Mean = 35.38 years

The mode (36.82) is slightly higher than the mean (35.38), indicating that the most common age group of patients is slightly older than the average age of all patients. Both values suggest that most patients admitted are in their mid-30s to mid-40s.

Problem 2

The following data gives the information on the observed life times (in hours) of 225 electrical components:

Lifetimes (in hours) 0-20 20-40 40-60 60-80 80-100 100-120
Frequency 10 35 52 61 38 29

Determine the modal lifetimes of the components.

Solution:

The modal class is the class with the highest frequency. Here, the highest frequency is 61, which corresponds to the class 60-80.

Using the mode formula:

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where:
L = lower limit of modal class = 60
f₁ = frequency of modal class = 61
f₀ = frequency of class preceding modal class = 52
f₂ = frequency of class succeeding modal class = 38
h = class width = 20

Mode = 60 + [(61 – 52)/(2×61 – 52 – 38)] × 20
= 60 + [9/(122 – 90)] × 20
= 60 + (9/32) × 20
= 60 + 5.625 = 65.625

Modal lifetime of components = 65.63 hours

Problem 3

The following data gives the distribution of total monthly household expenditure of 200 families of Gummadidala village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Expenditure (in rupees) 1000-1500 1500-2000 2000-2500 2500-3000 3000-3500 3500-4000 4000-4500 4500-5000
Number of families 24 40 33 28 30 22 16 7

Solution:

Mode Calculation:

The modal class is the class with the highest frequency. Here, the highest frequency is 40, which corresponds to the class 1500-2000.

Using the mode formula:

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where:
L = lower limit of modal class = 1500
f₁ = frequency of modal class = 40
f₀ = frequency of class preceding modal class = 24
f₂ = frequency of class succeeding modal class = 33
h = class width = 500

Mode = 1500 + [(40 – 24)/(2×40 – 24 – 33)] × 500
= 1500 + [16/(80 – 57)] × 500
= 1500 + (16/23) × 500
= 1500 + 347.83 ≈ 1847.83

Mean Calculation:

We’ll use the step-deviation method with assumed mean a = 2750 and class width h = 500

Class Interval Midpoint (xi) Frequency (fi) di = xi – a ui = di/h fiui
1000-1500 1250 24 -1500 -3 -72
1500-2000 1750 40 -1000 -2 -80
2000-2500 2250 33 -500 -1 -33
2500-3000 2750 (a) 28 0 0 0
3000-3500 3250 30 500 1 30
3500-4000 3750 22 1000 2 44
4000-4500 4250 16 1500 3 48
4500-5000 4750 7 2000 4 28
Total -35

Mean = \(a + \left(\frac{\sum f_iu_i}{\sum f_i}\right) \times h = 2750 + \left(\frac{-35}{200}\right) \times 500 = 2750 – 87.5 = 2662.5\)

Modal monthly expenditure = ₹1847.83
Mean monthly expenditure = ₹2662.50

Problem 4

The following distribution gives the state-wise, teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Number of students per teacher 15-20 20-25 25-30 30-35 35-40 40-45 45-50 50-55
Number of States 3 8 9 10 3 0 0 2

Solution:

Mode Calculation:

The modal class is the class with the highest frequency. Here, the highest frequency is 10, which corresponds to the class 30-35.

Using the mode formula:

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where:
L = lower limit of modal class = 30
f₁ = frequency of modal class = 10
f₀ = frequency of class preceding modal class = 9
f₂ = frequency of class succeeding modal class = 3
h = class width = 5

Mode = 30 + [(10 – 9)/(2×10 – 9 – 3)] × 5
= 30 + [1/(20 – 12)] × 5
= 30 + (1/8) × 5
= 30 + 0.625 = 30.625

Mean Calculation:

Class Interval Midpoint (xi) Frequency (fi) fixi
15-20 17.5 3 52.5
20-25 22.5 8 180
25-30 27.5 9 247.5
30-35 32.5 10 325
35-40 37.5 3 112.5
40-45 42.5 0 0
45-50 47.5 0 0
50-55 52.5 2 105
Total 35 1022.5

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{1022.5}{35} \approx 29.21\)

Interpretation:

Mode = 30.62 students per teacher
Mean = 29.21 students per teacher

The mode (30.62) is slightly higher than the mean (29.21), indicating that the most common student-teacher ratio is slightly higher than the average ratio across all states. This suggests that while most states have about 30-31 students per teacher, some states with lower ratios bring the average down slightly.

Problem 5

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs 3000-4000 4000-5000 5000-6000 6000-7000 7000-8000 8000-9000 9000-10000 10000-11000
Number of batsmen 4 18 9 7 6 3 1 1

Find the mode of the data.

Solution:

The modal class is the class with the highest frequency. Here, the highest frequency is 18, which corresponds to the class 4000-5000.

Using the mode formula:

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where:
L = lower limit of modal class = 4000
f₁ = frequency of modal class = 18
f₀ = frequency of class preceding modal class = 4
f₂ = frequency of class succeeding modal class = 9
h = class width = 1000

Mode = 4000 + [(18 – 4)/(2×18 – 4 – 9)] × 1000
= 4000 + [14/(36 – 13)] × 1000
= 4000 + (14/23) × 1000
= 4000 + 608.70 ≈ 4608.70

Modal runs scored by batsmen = 4608.70 runs

Problem 6

A student noted the number of cars passing through a spot on a road for 100 periods, each of 3 minutes, and summarised this in the table given below.

Number of cars 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency 7 14 13 12 20 11 15 8

Find the mode of the data.

Solution:

The modal class is the class with the highest frequency. Here, the highest frequency is 20, which corresponds to the class 40-50.

Using the mode formula:

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where:
L = lower limit of modal class = 40
f₁ = frequency of modal class = 20
f₀ = frequency of class preceding modal class = 12
f₂ = frequency of class succeeding modal class = 11
h = class width = 10

Mode = 40 + [(20 – 12)/(2×20 – 12 – 11)] × 10
= 40 + [8/(40 – 23)] × 10
= 40 + (8/17) × 10
= 40 + 4.71 ≈ 44.71

Mode number of cars passing = 44.71 cars per 3-minute period

10th Maths Statistics Exercise 14.1 Solutions

Exercise 14.1 Solutions – Class X Mathematics

Exercise 14.1 Solutions

Class X Mathematics
State Council of Educational Research and Training, Telangana, Hyderabad

Problem 1

A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of plants 0 – 2 2 – 4 4 – 6 6 – 8 8 – 10 10 – 12 12 – 14
Number of houses 1 2 1 5 6 2 3

Solution:

To find the mean, we’ll use the midpoint of each class interval and multiply by frequency.

Class Interval Midpoint (xi) Frequency (fi) fixi
0-2 1 1 1
2-4 3 2 6
4-6 5 1 5
6-8 7 5 35
8-10 9 6 54
10-12 11 2 22
12-14 13 3 39
Total 20 162

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{162}{20} = 8.1\)

Mean number of plants per house = 8.1

Problem 2

Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages in Rupees 200 – 250 250 – 300 300 – 350 350 – 400 400 – 450
Number of workers 12 14 8 6 10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Solution:

We’ll use the step-deviation method with assumed mean a = 325 and class width h = 50

Class Interval Midpoint (xi) Frequency (fi) di = xi – a ui = di/h fiui
200-250 225 12 -100 -2 -24
250-300 275 14 -50 -1 -14
300-350 325 (a) 8 0 0 0
350-400 375 6 50 1 6
400-450 425 10 100 2 20
Total -12

Mean = \(a + \left(\frac{\sum f_iu_i}{\sum f_i}\right) \times h = 325 + \left(\frac{-12}{50}\right) \times 50 = 325 – 12 = 313\)

Mean daily wages = ₹313

Problem 3

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹18. Find the missing frequency \( f \).

Daily pocket allowance (in Rupees) 11 – 13 13 – 15 15 – 17 17 – 19 19 – 21 21 – 23 23 – 25
Number of children 7 6 9 13 \( f \) 5 4

Solution:

Let’s calculate the sum of frequencies and products of midpoints and frequencies.

Class Interval Midpoint (xi) Frequency (fi) fixi
11-13 12 7 84
13-15 14 6 84
15-17 16 9 144
17-19 18 13 234
19-21 20 f 20f
21-23 22 5 110
23-25 24 4 96
Total 44 + f 752 + 20f

Given mean = 18

\(\frac{752 + 20f}{44 + f} = 18\)

752 + 20f = 792 + 18f

2f = 40

f = 20

The missing frequency \( f = 20 \)

Problem 4

Thirty women were examined in a hospital by a doctor and their heart beats per minute were recorded and summarised as shown. Find the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats/minute 65-68 68-71 71-74 74-77 77-80 80-83 83-86
Number of women 2 4 3 8 7 4 2

Solution:

We’ll use the direct method to find the mean.

Class Interval Midpoint (xi) Frequency (fi) fixi
65-68 66.5 2 133
68-71 69.5 4 278
71-74 72.5 3 217.5
74-77 75.5 8 604
77-80 78.5 7 549.5
80-83 81.5 4 326
83-86 84.5 2 169
Total 30 2277

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{2277}{30} = 75.9\)

Mean heart beats per minute = 75.9

Problem 5

In a retail market, fruit vendors were selling oranges kept in packing baskets. These baskets contained varying number of oranges. The following was the distribution of oranges.

Number of oranges 10-14 15-19 20-24 25-29 30-34
Number of baskets 15 110 135 115 25

Find the mean number of oranges kept in each basket. Which method of finding the mean did you choose?

Solution:

We’ll use the step-deviation method with assumed mean a = 22 and class width h = 5

Class Interval Midpoint (xi) Frequency (fi) di = xi – a ui = di/h fiui
10-14 12 15 -10 -2 -30
15-19 17 110 -5 -1 -110
20-24 22 (a) 135 0 0 0
25-29 27 115 5 1 115
30-34 32 25 10 2 50
Total 25

Mean = \(a + \left(\frac{\sum f_iu_i}{\sum f_i}\right) \times h = 22 + \left(\frac{25}{400}\right) \times 5 = 22 + \frac{125}{400} = 22 + 0.3125 = 22.3125\)

Mean number of oranges per basket = 22.31 (approx)

We chose the step-deviation method because the class intervals are equal and the frequencies are large.

Problem 6

The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in Rupees) 100-150 150-200 200-250 250-300 300-350
Number of households 4 5 12 2 2

Find the mean daily expenditure on food by a suitable method.

Solution:

We’ll use the direct method to find the mean.

Class Interval Midpoint (xi) Frequency (fi) fixi
100-150 125 4 500
150-200 175 5 875
200-250 225 12 2700
250-300 275 2 550
300-350 325 2 650
Total 25 5275

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{5275}{25} = 211\)

Mean daily expenditure on food = ₹211

Problem 7

To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Concentration of SO2 in ppm 0.00-0.04 0.04-0.08 0.08-0.12 0.12-0.16 0.16-0.20 0.20-0.24
Frequency 4 9 9 2 4 2

Find the mean concentration of SO2 in the air.

Solution:

We’ll use the direct method to find the mean.

Class Interval Midpoint (xi) Frequency (fi) fixi
0.00-0.04 0.02 4 0.08
0.04-0.08 0.06 9 0.54
0.08-0.12 0.10 9 0.90
0.12-0.16 0.14 2 0.28
0.16-0.20 0.18 4 0.72
0.20-0.24 0.22 2 0.44
Total 30 2.96

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{2.96}{30} \approx 0.0987\)

Mean concentration of SO2 = 0.099 ppm (approx)

Problem 8

A class teacher has the following attendance record of 40 students of a class for the whole term. Find the mean number of days a student was present out of 56 days in the term.

Number of days 35-38 38-41 41-44 44-47 47-50 50-53 53-56
Number of students 1 3 4 4 7 10 11

Solution:

We’ll use the direct method to find the mean.

Class Interval Midpoint (xi) Frequency (fi) fixi
35-38 36.5 1 36.5
38-41 39.5 3 118.5
41-44 42.5 4 170
44-47 45.5 4 182
47-50 48.5 7 339.5
50-53 51.5 10 515
53-56 54.5 11 599.5
Total 40 1961

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{1961}{40} = 49.025\)

Mean number of days present = 49.03 days (approx)

Problem 9

The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy rate in % 45-55 55-65 65-75 75-85 85-95
Number of cities 3 10 11 8 3

Solution:

We’ll use the direct method to find the mean.

Class Interval Midpoint (xi) Frequency (fi) fixi
45-55 50 3 150
55-65 60 10 600
65-75 70 11 770
75-85 80 8 640
85-95 90 3 270
Total 35 2430

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{2430}{35} \approx 69.43\)

Mean literacy rate = 69.43%

10th Maths Probability Exercise 13.2 Solutions

Exercise 13.2 Solutions – Class X Mathematics

Exercise 13.2 Solutions

Probability – Class X Mathematics

Question 1

A bag contains 3 red balls and 5 black balls. A ball is selected at random from the bag. What is the probability that the ball selected is:

(i) red? (ii) not red?

Solution:

Total number of balls = 3 (red) + 5 (black) = 8

(i) Probability of selecting a red ball = \(\frac{\text{Number of red balls}}{\text{Total balls}} = \frac{3}{8}\)

(ii) Probability of not selecting a red ball = 1 – Probability of selecting red ball = \(1 – \frac{3}{8} = \frac{5}{8}\)

Question 2

A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be:

(i) red? (ii) white? (iii) not green?

Solution:

Total marbles = 5 (red) + 8 (white) + 4 (green) = 17

(i) Probability of red marble = \(\frac{5}{17}\)

(ii) Probability of white marble = \(\frac{8}{17}\)

(iii) Probability of not green marble = 1 – Probability of green marble = \(1 – \frac{4}{17} = \frac{13}{17}\)

Question 3

A Kiddy bank contains hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin:

(i) will be a 50p coin? (ii) will not be a ₹5 coin?

Solution:

Total coins = 100 (50p) + 50 (₹1) + 20 (₹2) + 10 (₹5) = 180

(i) Probability of 50p coin = \(\frac{100}{180} = \frac{5}{9}\)

(ii) Probability of not ₹5 coin = 1 – Probability of ₹5 coin = \(1 – \frac{10}{180} = \frac{170}{180} = \frac{17}{18}\)

Question 4

Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?

Diagram Description: A rectangular fish tank containing 5 male fish (shown as smaller, colorful fish) and 8 female fish (shown as larger, less colorful fish). The shopkeeper is using a net to catch one fish at random.

Solution:

Total fish = 5 (male) + 8 (female) = 13

Probability of male fish = \(\frac{5}{13}\)

Question 5

A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (See figure), and these are equally likely outcomes. What is the probability that it will point at:

(i) 8? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9?

Diagram Description: A circular spinner divided into 8 equal sectors numbered 1 through 8 clockwise. An arrow is fixed at the center that can spin freely.

Solution:

Total possible outcomes = 8

(i) Probability of pointing at 8 = \(\frac{1}{8}\)

(ii) Odd numbers = {1, 3, 5, 7} → 4 outcomes. Probability = \(\frac{4}{8} = \frac{1}{2}\)

(iii) Numbers > 2 = {3, 4, 5, 6, 7, 8} → 6 outcomes. Probability = \(\frac{6}{8} = \frac{3}{4}\)

(iv) Numbers < 9 = {1, 2, 3, 4, 5, 6, 7, 8} → 8 outcomes. Probability = \(\frac{8}{8} = 1\)

Question 6

One card is selected from a well-shuffled deck of 52 cards. Find the probability of getting:

(i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds

Solution:

Total cards = 52

(i) King of red colour: There are 2 (King of Hearts and King of Diamonds). Probability = \(\frac{2}{52} = \frac{1}{26}\)

(ii) Face cards: Jack, Queen, King in each suit → 3 × 4 = 12. Probability = \(\frac{12}{52} = \frac{3}{13}\)

(iii) Red face cards: Face cards in hearts and diamonds → 3 × 2 = 6. Probability = \(\frac{6}{52} = \frac{3}{26}\)

(iv) Jack of hearts: Only 1 card. Probability = \(\frac{1}{52}\)

(v) Spade cards: 13. Probability = \(\frac{13}{52} = \frac{1}{4}\)

(vi) Queen of diamonds: Only 1 card. Probability = \(\frac{1}{52}\)

Question 7

Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is selected at random.

(i) What is the probability that the card is the queen?

(ii) If the queen is selected and put aside (without replacement), what is the probability that the second card selected is (a) an ace? (b) a queen?

Solution:

Total cards initially = 5

(i) Probability of queen = \(\frac{1}{5}\)

(ii) After removing queen, remaining cards = 4

(a) Probability of ace = \(\frac{1}{4}\) (only ace of diamonds left)

(b) Probability of queen = \(\frac{0}{4} = 0\) (queen has been removed)

Question 8

12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Solution:

Total pens = 12 (defective) + 132 (good) = 144

Probability of good pen = \(\frac{132}{144} = \frac{11}{12}\)

Question 9

A lot of 20 bulbs contain 4 defective ones. One bulb is selected at random from the lot. What is the probability that this bulb is defective? Suppose the bulb selected in previous case is not defective and is not replaced. Now one bulb is selected at random from the rest. What is the probability that this bulb is not defective?

Solution:

First selection:

Total bulbs = 20, Defective = 4

Probability of defective bulb = \(\frac{4}{20} = \frac{1}{5}\)

Second selection (after removing one good bulb):

Remaining bulbs = 19, Non-defective = 16 – 1 = 15

Probability of not defective bulb = \(\frac{15}{19}\)

Question 10

A box contains 90 discs which are numbered from 1 to 90. If one disc is selected at random from the box, find the probability that it bears:

(i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.

Solution:

Total discs = 90

(i) Two-digit numbers: 10 to 90 → 81 numbers. Probability = \(\frac{81}{90} = \frac{9}{10}\)

(ii) Perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81 → 9 numbers. Probability = \(\frac{9}{90} = \frac{1}{10}\)

(iii) Divisible by 5: 5, 10, 15, …, 90 → 18 numbers. Probability = \(\frac{18}{90} = \frac{1}{5}\)

Question 11

Suppose you drop a die at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with diameter 1m?

Diagram Description: A rectangular region with length 3m and width 2m. Inside it, there’s a circle with diameter 1m (radius 0.5m) centered in the rectangle.

Solution:

Area of rectangle = length × width = 3 × 2 = 6 m²

Area of circle = πr² = π(0.5)² = 0.25π m²

Probability = \(\frac{\text{Area of circle}}{\text{Area of rectangle}} = \frac{0.25π}{6} = \frac{π}{24}\)

Question 12

A lot consists of 144 ball pens of which 20 are defective and the others are good. The shopkeeper draws one pen at random and gives it to Sudha. What is the probability that:

(i) She will buy it? (ii) She will not buy it?

Solution:

Assuming Sudha will buy if the pen is good:

Good pens = 144 – 20 = 124

(i) Probability she will buy = Probability of good pen = \(\frac{124}{144} = \frac{31}{36}\)

(ii) Probability she will not buy = Probability of defective pen = \(\frac{20}{144} = \frac{5}{36}\)

Question 13

Two dice are rolled simultaneously and counts are added:

(i) complete the table given below:

Event: ‘Sum on 2 dice’ 2 3 4 5 6 7 8 9 10 11 12
Probability \(\frac{1}{36}\) \(\frac{2}{36}\) \(\frac{3}{36}\) \(\frac{4}{36}\) \(\frac{5}{36}\) \(\frac{6}{36}\) \(\frac{5}{36}\) \(\frac{4}{36}\) \(\frac{3}{36}\) \(\frac{2}{36}\) \(\frac{1}{36}\)

(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability \(\frac{1}{11}\). Do you agree with this argument? Justify your answer.

Solution:

(i) Table completed above.

(ii) No, I don’t agree. While there are 11 possible sums, they are not equally likely. For example, there’s only one way to get a sum of 2 (1+1) but six ways to get a sum of 7 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1). Therefore, the probability of each sum is different.

Question 14

A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Deskhtha wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that she will lose the game.

Solution:

Total possible outcomes when tossing a coin 3 times = 2³ = 8

Favorable outcomes for winning: HHH, TTT → 2 outcomes

Probability of winning = \(\frac{2}{8} = \frac{1}{4}\)

Probability of losing = 1 – Probability of winning = \(1 – \frac{1}{4} = \frac{3}{4}\)

Question 15

A dice is thrown twice. What is the probability that:

(i) 5 will not come up either time? (ii) 5 will come up at least once?

Solution:

Total possible outcomes when throwing a die twice = 6 × 6 = 36

(i) Outcomes where 5 doesn’t appear either time: Each die has 5 options (1,2,3,4,6). Total = 5 × 5 = 25

Probability = \(\frac{25}{36}\)

(ii) Probability that 5 comes at least once = 1 – Probability that 5 doesn’t appear at all = \(1 – \frac{25}{36} = \frac{11}{36}\)

10th Maths Probability Exercise 13.1 Solutions

Exercise 13.1 Solutions – Class X Mathematics

Exercise 13.1 Solutions

Probability – Class X Mathematics

State Council of Educational Research and Training, Telangana, Hyderabad

Question 1: Complete the following statements

(i) Probability of an event \( E + \text{Probability of the event ‘not } E’ = \) ______.

Solution: The sum of the probability of an event and its complement is always 1.

\[ P(E) + P(\text{not } E) = 1 \]

(ii) The probability of an event that cannot happen is ______. Such an event is called ______.

Solution: The probability of an impossible event is 0. Such an event is called an impossible event.

\[ P(\text{Impossible event}) = 0 \]

(iii) The probability of an event that is certain to happen is ______. Such an event is called ______.

Solution: The probability of a certain event is 1. Such an event is called a sure event.

\[ P(\text{Certain event}) = 1 \]

(iv) The sum of the probabilities of all the elementary events of an experiment is ______.

Solution: The sum of probabilities of all elementary events of an experiment is 1.

\[ \sum P(E_i) = 1 \]

(v) The probability of an event is greater than or equal to ______ and less than or equal to ______.

Solution: The probability of any event \( E \) satisfies:

\[ 0 \leq P(E) \leq 1 \]

Question 2: Which of the following experiments have equally likely outcomes? Explain.

(i) A driver attempts to start a car. The car starts or does not start.

Solution: Not equally likely. The probability depends on the car’s condition, fuel, etc.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.

Solution: Not equally likely. The outcome depends on the player’s skill.

(iii) A trial is made to answer a true-false question. The answer is right or wrong.

Solution: Equally likely. There’s a 50% chance of guessing correctly.

\[ P(\text{Right}) = P(\text{Wrong}) = 0.5 \]

(iv) A baby is born. It is a boy or a girl.

Solution: Equally likely (assuming equal probability for biological sexes).

\[ P(\text{Boy}) = P(\text{Girl}) ≈ 0.5 \]

Question 3: If \( P(E) = 0.05 \), what is the probability of ‘not E’?

Solution:

We know that \( P(E) + P(\text{not } E) = 1 \)

\[ P(\text{not } E) = 1 – P(E) = 1 – 0.05 = 0.95 \]

Question 4: A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag.

(i) an orange flavoured candy?

Solution: Since the bag contains only lemon flavored candies:

\[ P(\text{Orange}) = 0 \] (Impossible event)

(ii) a lemon flavoured candy?

Solution: Since all candies are lemon flavored:

\[ P(\text{Lemon}) = 1 \] (Certain event)

Diagram: A bag containing multiple lemon candies (all yellow) with one being drawn out.

[Illustration of a bag with yellow candies and one hand taking a candy out]

Question 5: Rahim removes all the hearts from the cards. What is the probability of:

i. Getting an ace from the remaining pack.

Solution: Original deck has 52 cards. After removing 13 hearts, 39 cards remain.

Number of aces in remaining cards: 3 (since Ace of Hearts was removed)

\[ P(\text{Ace}) = \frac{3}{39} = \frac{1}{13} \]

ii. Getting a diamonds.

Solution: All 13 diamonds remain in the deck of 39 cards.

\[ P(\text{Diamond}) = \frac{13}{39} = \frac{1}{3} \]

iii. Getting a card that is not a heart.

Solution: Since all hearts have been removed, all remaining 39 cards are not hearts.

\[ P(\text{Not heart}) = \frac{39}{39} = 1 \]

iv. Getting the Ace of hearts.

Solution: Since all hearts have been removed:

\[ P(\text{Ace of Hearts}) = 0 \] (Impossible event)

Diagram: A standard deck of cards with all heart cards (red) removed, showing only clubs, diamonds, and spades remaining.

[Illustration of a partial deck with hearts missing]

Question 6: In a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Solution:

Let \( P(\text{Same}) \) be the probability that two students share a birthday.

Given \( P(\text{Not same}) = 0.992 \)

\[ P(\text{Same}) = 1 – P(\text{Not same}) = 1 – 0.992 = 0.008 \]

Question 7: A die is rolled once. Find the probability of getting:

(i) a prime number

Solution: Prime numbers on a die: 2, 3, 5 (3 outcomes)

Total possible outcomes: 6

\[ P(\text{Prime}) = \frac{3}{6} = \frac{1}{2} \]

(ii) a number lying between 2 and 6

Solution: Numbers between 2 and 6: 3, 4, 5 (3 outcomes)

\[ P(\text{Between 2 and 6}) = \frac{3}{6} = \frac{1}{2} \]

(iii) an odd number

Solution: Odd numbers on a die: 1, 3, 5 (3 outcomes)

\[ P(\text{Odd}) = \frac{3}{6} = \frac{1}{2} \]

Diagram: A standard six-faced die showing numbers 1 through 6 with prime numbers (2,3,5) highlighted.

[Illustration of a die with some numbers highlighted]

Question 8: What is the probability of selecting a red king from a deck of cards?

Solution:

Total cards in deck: 52

Red kings: King of Hearts and King of Diamonds (2 cards)

\[ P(\text{Red King}) = \frac{2}{52} = \frac{1}{26} \]

Diagram: A deck of cards showing the two red kings (King of Hearts and King of Diamonds).

[Illustration of two red king cards]

Question 9: Make 5 more problems of this kind using dice, cards or birthdays

Sample Problems:

  1. What is the probability of rolling a number greater than 4 on a standard die?
  2. If two cards are drawn from a standard deck without replacement, what is the probability both are spades?
  3. In a group of 30 people, what is the probability that at least two people share the same birthday?
  4. What is the probability of drawing a face card (Jack, Queen, King) from a standard deck?
  5. If you roll two dice, what is the probability that the sum is 7?

These problems can be discussed with friends and teachers to understand probability concepts better.

10th Maths Applications of Trigonometry Exercise 12.2 Solutions

Exercise 11.3 Solutions - Class X Mathematics

Exercise 11.3 Solutions

Some Applications of Trigonometry

Class X Mathematics Textbook
State Council of Educational Research and Training, Telangana, Hyderabad

Problem 1

A TV tower stands vertically on the side of a road. From a point on the other side directly opposite to the tower, the angle of elevation of the top of tower is \( 60^\circ \). From another point 10 m away from this point, on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is \( 30^\circ \). Find the height of the tower and the width of the road.

Diagram Description: A vertical TV tower on one side of a road. Point A is directly opposite the tower on the other side of the road, with angle of elevation 60°. Point B is 10m from A towards the tower, with angle of elevation 30°.
Road 60° 30° Tower (h) Point A Point B 10 m Width (x)

Solution:

Let the height of the tower be \( h \) meters and the width of the road be \( x \) meters.

From point A (directly opposite):

\[ \tan 60^\circ = \frac{h}{x} \Rightarrow h = x\sqrt{3} \]

From point B (10m from A towards the tower):

\[ \tan 30^\circ = \frac{h}{x + 10} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x + 10} \]

Substitute \( h = x\sqrt{3} \):

\[ \frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{x + 10} \]

\[ x + 10 = 3x \]

\[ 2x = 10 \Rightarrow x = 5 \text{ meters} \]

Then \( h = 5\sqrt{3} \) meters.

The height of the tower is \( 5\sqrt{3} \) meters and the width of the road is 5 meters.

Problem 2

A 1.5 m tall boy is looking at the top of a temple which is 30 meter in height from a point at certain distance. The angle of elevation from his eye to the top of the crown of the temple increases from 30° to 60° as he walks towards the temple. Find the distance he walked towards the temple.

Diagram Description: A temple of height 30m. A boy of height 1.5m stands at point A (initial position) with angle of elevation 30°, then moves to point B (closer position) with angle of elevation 60°.
30 m 1.5 m 30° 60° Point A Point B Distance walked (d) 28.5 m

Solution:

Effective height of temple above boy's eye level = 30 - 1.5 = 28.5 meters

Let initial distance be \( x \) meters and distance walked be \( d \) meters.

From initial position:

\[ \tan 30^\circ = \frac{28.5}{x} \Rightarrow x = 28.5\sqrt{3} \]

From final position:

\[ \tan 60^\circ = \frac{28.5}{x - d} \Rightarrow x - d = \frac{28.5}{\sqrt{3}} \]

Substitute \( x = 28.5\sqrt{3} \):

\[ 28.5\sqrt{3} - d = \frac{28.5}{\sqrt{3}} \]

\[ d = 28.5\sqrt{3} - \frac{28.5}{\sqrt{3}} \]

\[ d = 28.5\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) = 28.5\left(\frac{3 - 1}{\sqrt{3}}\right) = \frac{57}{\sqrt{3}} = 19\sqrt{3} \text{ meters} \]

The boy walked \( 19\sqrt{3} \) meters towards the temple.

Problem 3

A statue stands on the top of a 2m tall pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the statue.

Diagram Description: A pedestal of height 2m with a statue on top. From a point on the ground, angle to top of pedestal is 45° and angle to top of statue is 60°.
45° 60° Pedestal (2m) Statue (h) Observation point Distance (x)

Solution:

Let the height of the statue be \( h \) meters and distance from point to pedestal be \( x \) meters.

For pedestal (2m height):

\[ \tan 45^\circ = \frac{2}{x} \Rightarrow x = 2 \text{ meters} \]

For statue (2 + h meters height):

\[ \tan 60^\circ = \frac{2 + h}{2} \Rightarrow \sqrt{3} = \frac{2 + h}{2} \]

\[ 2 + h = 2\sqrt{3} \Rightarrow h = 2\sqrt{3} - 2 = 2(\sqrt{3} - 1) \text{ meters} \]

The height of the statue is \( 2(\sqrt{3} - 1) \) meters.

Problem 4

From the top of a building, the angle of elevation of the top of a cell tower is 60° and the angle of depression to its foot is 45°. If distance of the building from the tower is 7m, then find the height of the tower.

Diagram Description: A building and a cell tower 7m apart. From the top of the building, angle of elevation to tower top is 60° and angle of depression to tower base is 45°.
45° 60° Building (h) Tower (H) 7 m

Solution:

Let height of building be \( h \) meters and height of tower be \( H \) meters.

From angle of depression (45°):

\[ \tan 45^\circ = \frac{h}{7} \Rightarrow h = 7 \text{ meters} \]

From angle of elevation (60°):

\[ \tan 60^\circ = \frac{H - h}{7} \Rightarrow \sqrt{3} = \frac{H - 7}{7} \]

\[ H - 7 = 7\sqrt{3} \Rightarrow H = 7 + 7\sqrt{3} = 7(1 + \sqrt{3}) \text{ meters} \]

The height of the tower is \( 7(1 + \sqrt{3}) \) meters.

Problem 5

A wire of length 18 m had been tied with electric pole at an angle of elevation 30° with the ground. Because it was covering a long distance, it was cut and tied at an angle of elevation 60° with the ground. How much length of the wire was cut?

Diagram Description: An electric pole with two positions of a wire - original at 30° (18m) and shortened at 60°.
30° 60° Pole (h) Original wire (18m) New wire

Solution:

Let height of pole be \( h \) meters.

Original wire (18m at 30°):

\[ \sin 30^\circ = \frac{h}{18} \Rightarrow h = 18 \times \frac{1}{2} = 9 \text{ meters} \]

New wire (at 60°):

\[ \sin 60^\circ = \frac{9}{L} \Rightarrow L = \frac{9}{\sin 60^\circ} = \frac{9}{\sqrt{3}/2} = \frac{18}{\sqrt{3}} = 6\sqrt{3} \text{ meters} \]

Length cut = Original - New = \( 18 - 6\sqrt{3} = 6(3 - \sqrt{3}) \) meters

\( 6(3 - \sqrt{3}) \) meters of wire was cut.

Problem 6

The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 30 m high, find the height of the building.

Diagram Description: A tower and building separated by some distance. From tower base to building top is 30°, from building base to tower top is 60°.
30° 60° Tower (30m) Building (h) Distance (d)

Solution:

Let height of building be \( h \) meters and distance between them be \( d \) meters.

From building's foot to tower's top (30m):

\[ \tan 60^\circ = \frac{30}{d} \Rightarrow d = \frac{30}{\sqrt{3}} = 10\sqrt{3} \text{ meters} \]

From tower's foot to building's top:

\[ \tan 30^\circ = \frac{h}{d} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{10\sqrt{3}} \]

\[ h = \frac{10\sqrt{3}}{\sqrt{3}} = 10 \text{ meters} \]

The height of the building is 10 meters.

Problem 7

Two poles of equal heights are standing opposite to each other on either side of the road, which is 120 feet wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles.

Diagram Description: Two poles of equal height on opposite sides of a 120ft road. A point between them has angles of elevation 60° to one pole and 30° to the other.
120 feet 60° 30° Pole (h) Pole (h) x 120-x Point

Solution:

Let height of poles be \( h \) feet.

Let distances from point to poles be \( x \) and \( 120 - x \) feet.

For first pole (60°):

\[ \tan 60^\circ = \frac{h}{x} \Rightarrow h = x\sqrt{3} \]

For second pole (30°):

\[ \tan 30^\circ = \frac{h}{120 - x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{120 - x} \]

Substitute \( h = x\sqrt{3} \):

\[ \frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{120 - x} \]

\[ 120 - x = 3x \]

\[ 4x = 120 \Rightarrow x = 30 \text{ feet} \]

Then \( h = 30\sqrt{3} \) feet

Distances: 30 feet and 90 feet

The poles are \( 30\sqrt{3} \) feet high. The point is 30 feet from one pole and 90 feet from the other.

Problem 8

The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Find the height of the tower.

Diagram Description: A tower with two observation points at 4m and 9m from base. Their angles of elevation are complementary (sum to 90°).
θ 90°-θ Tower (h) 4 m 5 m 9 m

Solution:

Let height of tower be \( h \) meters.

Let angles be \( \theta \) and \( 90^\circ - \theta \).

From first point (4m):

\[ \tan \theta = \frac{h}{4} \]

From second point (9m):

\[ \tan(90^\circ - \theta) = \frac{h}{9} \Rightarrow \cot \theta = \frac{h}{9} \]

Since \( \tan \theta \times \cot \theta = 1 \):

\[ \frac{h}{4} \times \frac{h}{9} = 1 \]

\[ h^2 = 36 \Rightarrow h = 6 \text{ meters} \]

The height of the tower is 6 meters.

Problem 9

The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of \( 1500\sqrt{3} \) meter, find the speed of the jet plane.

Diagram Description: A jet flying at constant height. From point A, initial angle is 60°, after 15 seconds angle is 30°.
60° 30° Point A Initial position After 15 sec \(1500\sqrt{3}\) m Distance flown (d)

Solution:

Let initial distance be \( x \) meters and height \( h = 1500\sqrt{3} \) meters.

Initial position (60°):

\[ \tan 60^\circ = \frac{h}{x} \Rightarrow x = \frac{1500\sqrt{3}}{\sqrt{3}} = 1500 \text{ meters} \]

After 15 seconds (30°):

\[ \tan 30^\circ = \frac{h}{x + d} \Rightarrow \frac{1}{\sqrt{3}} = \frac{1500\sqrt{3}}{1500 + d} \]

\[ 1500 + d = 1500 \times 3 = 4500 \]

\[ d = 3000 \text{ meters} \]

Speed = Distance/Time = \( \frac{3000}{15} = 200 \) m/s

Convert to km/h: \( 200 \times \frac{18}{5} = 720 \) km/h

The speed of the jet plane is 720 km/h.

Problem 10

The angle of elevation of the top of a tower from the foot of the building is 30° and the angle of elevation of the top of the building from the foot of the tower is 60°. What is the ratio of heights of tower and building.

Diagram Description: A tower and building separated by some distance. From building base to tower top is 30°, from tower base to building top is 60°.
30° 60° Tower (h_t) Building (h_b) Distance (d)

Solution:

Let height of tower be \( h_t \) and building be \( h_b \), distance between them be \( d \).

From building's foot to tower's top:

\[ \tan 30^\circ = \frac{h_t}{d} \Rightarrow d = h_t\sqrt{3} \]

From tower's foot to building's top:

\[ \tan 60^\circ = \frac{h_b}{d} \Rightarrow \sqrt{3} = \frac{h_b}{h_t\sqrt{3}} \]

\[ h_b = 3h_t \]

Ratio \( \frac{h_t}{h_b} = \frac{h_t}{3h_t} = \frac{1}{3} \)

The ratio of heights of tower to building is 1:3.

Class X Mathematics - SCERT Telangana

10th Maths Applications of Trigonometry Exercise 12.1 Solutions

Exercise 12.1 Solutions - Class X Mathematics

Exercise 12.1 Solutions

Class X Mathematics - State Council of Educational Research and Training, Telangana, Hyderabad

Problem 1

A tower stands vertically on the ground. From a point which is 15 meter away from the foot of the tower, the angle of elevation of the top of the tower is \(45^\circ\). What is the height of the tower?

45° Tower 15 m h Observer

Right triangle with tower height h, ground distance 15m, and angle of elevation 45°

Solution:

Let the height of the tower be \( h \) meters.

Given: Distance from tower = 15 m, Angle of elevation = \( 45^\circ \)

Using trigonometric ratio:

\[ \tan 45^\circ = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{h}{15} \]

\[ 1 = \frac{h}{15} \]

\[ h = 15 \text{ meters} \]

The height of the tower is 15 meters.

Problem 2

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground by making \(30^\circ\) angle with the ground. The distance between the foot of the tree and the top of the tree on the ground is 6m. Find the height of the tree before falling down.

30° h 6 m Standing part x (broken part)

Broken tree with standing height h, broken part length x, ground distance 6m, and angle 30°

Solution:

Let the height of the remaining part be \( h \) meters and the broken part be \( x \) meters.

Given: Distance on ground = 6 m, Angle = \( 30^\circ \)

Using trigonometric ratios:

\[ \cos 30^\circ = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{6}{x} \]

\[ \frac{\sqrt{3}}{2} = \frac{6}{x} \]

\[ x = \frac{12}{\sqrt{3}} = 4\sqrt{3} \text{ meters} \]

\[ \tan 30^\circ = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{h}{6} \]

\[ \frac{1}{\sqrt{3}} = \frac{h}{6} \]

\[ h = \frac{6}{\sqrt{3}} = 2\sqrt{3} \text{ meters} \]

Total height = \( h + x = 2\sqrt{3} + 4\sqrt{3} = 6\sqrt{3} \) meters

The original height of the tree was \( 6\sqrt{3} \) meters.

Problem 3

A contractor wants to set up a slide for the children to play in the park. He wants to set it up at the height of 2 m and by making an angle of \(30^\circ\) with the ground. What should be the length of the slide?

30° 2 m L (slide)

Right triangle with height 2m, slide length L, and angle 30° with ground

Solution:

Let the length of the slide be \( L \) meters.

Given: Height = 2 m, Angle = \( 30^\circ \)

Using trigonometric ratio:

\[ \sin 30^\circ = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{2}{L} \]

\[ \frac{1}{2} = \frac{2}{L} \]

\[ L = 4 \text{ meters} \]

The length of the slide should be 4 meters.

Problem 4

Length of the shadow of a 15 meter high pole is \(15\sqrt{3}\) meters at 8 O'clock in the morning. Then, what is the angle of elevation of the Sunrays with the ground at the time?

θ 15 m \(15\sqrt{3}\) m Sun rays

Right triangle with pole height 15m, shadow length 15√3m, and angle of elevation θ

Solution:

Given: Height of pole = 15 m, Shadow length = \( 15\sqrt{3} \) m

Let the angle of elevation be \( \theta \).

Using trigonometric ratio:

\[ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{15}{15\sqrt{3}} = \frac{1}{\sqrt{3}} \]

\[ \tan \theta = \frac{1}{\sqrt{3}} \]

\[ \theta = 30^\circ \]

The angle of elevation of the Sun is \( 30^\circ \).

Problem 5

You want to erect a pole of height 10 m with the support of three ropes. Each rope has to make an angle \(30^\circ\) with the pole. What should be the length of the rope?

30° 10 m L (rope)

Right triangle with pole height 10m, rope length L, and angle 30° with pole

Solution:

Let the length of each rope be \( L \) meters.

Given: Pole height = 10 m, Angle with pole = \( 30^\circ \)

Using trigonometric ratio:

\[ \cos 30^\circ = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{10}{L} \]

\[ \frac{\sqrt{3}}{2} = \frac{10}{L} \]

\[ L = \frac{20}{\sqrt{3}} = \frac{20\sqrt{3}}{3} \text{ meters} \]

Each rope should be \( \frac{20\sqrt{3}}{3} \) meters long.

Problem 6

Suppose you are shooting an arrow from the top of a building at an height of 6 m to a target on the ground at an angle of depression of 60°. What is the distance between you and the object?

60° 6 m Distance Target

Right triangle with building height 6m, angle of depression 60°

Solution:

Given: Building height = 6 m, Angle of depression = \( 60^\circ \)

Angle of elevation from target to person = \( 60^\circ \) (alternate angles)

Let the distance between building and target be \( d \) meters.

Using trigonometric ratio:

\[ \tan 60^\circ = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{6}{d} \]

\[ \sqrt{3} = \frac{6}{d} \]

\[ d = \frac{6}{\sqrt{3}} = 2\sqrt{3} \text{ meters} \]

The actual distance between you and the target is the hypotenuse:

\[ \text{Distance} = \frac{6}{\sin 60^\circ} = \frac{6}{\sqrt{3}/2} = \frac{12}{\sqrt{3}} = 4\sqrt{3} \text{ meters} \]

The distance between you and the object is \( 4\sqrt{3} \) meters.

Problem 7

An electrician wants to repair an electric connection on a pole of height 9 m. He needs to reach 1.8 m below the top of the pole to do repair work. What should be the length of the ladder which he should use, when he climbs it at an angle of 60° with the ground? What will be the distance between foot of the ladder and foot of the pole?

60° 7.2 m L (ladder) d

Right triangle with repair height 7.2m, ladder length L, and angle 60° with ground

Solution:

Given: Pole height = 9 m, Repair height = 9 - 1.8 = 7.2 m, Angle = \( 60^\circ \)

Let ladder length be \( L \) meters and ground distance be \( d \) meters.

Using trigonometric ratios:

\[ \sin 60^\circ = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{7.2}{L} \]

\[ \frac{\sqrt{3}}{2} = \frac{7.2}{L} \]

\[ L = \frac{14.4}{\sqrt{3}} = \frac{14.4\sqrt{3}}{3} = 4.8\sqrt{3} \text{ meters} \]

\[ \tan 60^\circ = \frac{7.2}{d} \]

\[ \sqrt{3} = \frac{7.2}{d} \]

\[ d = \frac{7.2}{\sqrt{3}} = 2.4\sqrt{3} \text{ meters} \]

The ladder should be \( 4.8\sqrt{3} \) meters long and its foot should be \( 2.4\sqrt{3} \) meters from the pole.

Problem 8

A boat has to cross a river. It crosses the river by making an angle of 60° with the bank of the river due to the stream of the river and travels a distance of 600m to reach the another side of the river. What is the width of the river?

60° 600 m w (width)

Right triangle with boat path 600m, river width w, and angle 60° with bank

Solution:

Let the width of the river be \( w \) meters.

Given: Diagonal distance = 600 m, Angle with bank = \( 60^\circ \)

Using trigonometric ratio:

\[ \sin 60^\circ = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{w}{600} \]

\[ \frac{\sqrt{3}}{2} = \frac{w}{600} \]

\[ w = 600 \times \frac{\sqrt{3}}{2} = 300\sqrt{3} \text{ meters} \]

The width of the river is \( 300\sqrt{3} \) meters.

Problem 9

An observer of height 1.8 m is 13.2 m away from a palm tree. The angle of elevation of the top of the tree from his eyes is 45°. What is the height of the palm tree?

45° 1.8 m 13.2 m h

Right triangle with observer height 1.8m, distance 13.2m, angle of elevation 45°

Solution:

Given: Observer height = 1.8 m, Distance = 13.2 m, Angle of elevation = \( 45^\circ \)

Let the height of the tree above observer's eyes be \( h \) meters.

Using trigonometric ratio:

\[ \tan 45^\circ = \frac{h}{13.2} \]

\[ 1 = \frac{h}{13.2} \]

\[ h = 13.2 \text{ meters} \]

Total tree height = Observer height + \( h \) = 1.8 + 13.2 = 15 meters

The height of the palm tree is 15 meters.

Problem 10

In the adjacent figure, AC = 6 cm, AB = 5 cm and \(\angle BAC\) = 30°. Find the area of the triangle.

30° AC = 6 cm AB = 5 cm A C B

Triangle ABC with sides AB=5cm, AC=6cm, and included angle 30°

Solution:

Given: AB = 5 cm, AC = 6 cm, \(\angle BAC\) = \( 30^\circ \)

Area of triangle when two sides and included angle are known:

\[ \text{Area} = \frac{1}{2} \times AB \times AC \times \sin \theta \]

\[ \text{Area} = \frac{1}{2} \times 5 \times 6 \times \sin 30^\circ \]

\[ \text{Area} = \frac{1}{2} \times 30 \times \frac{1}{2} = 7.5 \text{ cm}^2 \]

The area of the triangle is 7.5 cm².

10th Maths Trigonometry Exercise 11.4 Solutions

Exercise 11.4 Solutions

Exercise 11.4 Solutions

From Class X Mathematics textbook by the State Council of Educational Research and Training, Telangana, Hyderabad

1. Evaluate the following:
(i) \( (1 + \tan \theta + \sec \theta) (1 + \cot \theta – \csc \theta) \)

Let’s expand the expression:

\( = 1(1 + \cot \theta – \csc \theta) + \tan \theta(1 + \cot \theta – \csc \theta) + \sec \theta(1 + \cot \theta – \csc \theta) \)

\( = 1 + \cot \theta – \csc \theta + \tan \theta + \tan \theta \cot \theta – \tan \theta \csc \theta + \sec \theta + \sec \theta \cot \theta – \sec \theta \csc \theta \)

Simplify using trigonometric identities:

\( \tan \theta \cot \theta = 1 \)

\( \tan \theta \csc \theta = \frac{\sin \theta}{\cos \theta} \cdot \frac{1}{\sin \theta} = \frac{1}{\cos \theta} = \sec \theta \)

\( \sec \theta \cot \theta = \frac{1}{\cos \theta} \cdot \frac{\cos \theta}{\sin \theta} = \frac{1}{\sin \theta} = \csc \theta \)

Substituting back:

\( = 1 + \cot \theta – \csc \theta + \tan \theta + 1 – \sec \theta + \sec \theta + \csc \theta – \sec \theta \csc \theta \)

Many terms cancel out:

\( = 2 + \cot \theta + \tan \theta – \sec \theta \csc \theta \)

(ii) \( (\sin \theta + \cos \theta)^2 + (\sin \theta – \cos \theta)^2 \)

Expand both squares:

\( = (\sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta) + (\sin^2 \theta – 2\sin \theta \cos \theta + \cos^2 \theta) \)

Combine like terms:

\( = 2\sin^2 \theta + 2\cos^2 \theta \)

Factor out 2:

\( = 2(\sin^2 \theta + \cos^2 \theta) \)

Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):

\( = 2(1) = 2 \)

(iii) \( (\sec^2 \theta – 1) (\csc^2 \theta – 1) \)

We know that:

\( \sec^2 \theta – 1 = \tan^2 \theta \)

\( \csc^2 \theta – 1 = \cot^2 \theta \)

So the expression becomes:

\( \tan^2 \theta \cdot \cot^2 \theta \)

Since \( \cot \theta = \frac{1}{\tan \theta} \):

\( = \tan^2 \theta \cdot \frac{1}{\tan^2 \theta} = 1 \)

2. Show that \( (\csc \theta – \cot \theta)^2 = \frac{1 – \cos \theta}{1 + \cos \theta} \)

Start with the left side:

\( (\csc \theta – \cot \theta)^2 \)

Express in terms of sine and cosine:

\( = \left( \frac{1}{\sin \theta} – \frac{\cos \theta}{\sin \theta} \right)^2 = \left( \frac{1 – \cos \theta}{\sin \theta} \right)^2 = \frac{(1 – \cos \theta)^2}{\sin^2 \theta} \)

Using \( \sin^2 \theta = 1 – \cos^2 \theta \):

\( = \frac{(1 – \cos \theta)^2}{(1 – \cos \theta)(1 + \cos \theta)} \)

Cancel \( (1 – \cos \theta) \):

\( = \frac{1 – \cos \theta}{1 + \cos \theta} \)

Which matches the right side.

3. Show that \( \sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A \)

Start with the left side:

Rationalize the numerator by multiplying numerator and denominator by \( 1 + \sin A \):

\( \sqrt{\frac{(1 + \sin A)^2}{1 – \sin^2 A}} = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} = \frac{1 + \sin A}{\cos A} \)

Split the fraction:

\( = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A \)

Which matches the right side.

4. Show that \( \frac{1 – \tan^2 A}{\cot^2 A – 1} = \tan^2 A \)

Start with the left side:

Express everything in terms of tan A:

\( \cot A = \frac{1}{\tan A} \), so \( \cot^2 A = \frac{1}{\tan^2 A} \)

Substitute:

\( \frac{1 – \tan^2 A}{\frac{1}{\tan^2 A} – 1} = \frac{1 – \tan^2 A}{\frac{1 – \tan^2 A}{\tan^2 A}} \)

Divide by a fraction is same as multiplying by its reciprocal:

\( = (1 – \tan^2 A) \cdot \frac{\tan^2 A}{1 – \tan^2 A} = \tan^2 A \)

Which matches the right side.

5. Show that \( \frac{1}{\cos \theta} – \cos \theta = \tan \theta \cdot \sin \theta \)

Start with the left side:

\( \frac{1}{\cos \theta} – \cos \theta = \frac{1 – \cos^2 \theta}{\cos \theta} = \frac{\sin^2 \theta}{\cos \theta} \)

Now, the right side:

\( \tan \theta \cdot \sin \theta = \frac{\sin \theta}{\cos \theta} \cdot \sin \theta = \frac{\sin^2 \theta}{\cos \theta} \)

Both sides are equal.

6. Simplify \( \sec A (1 – \sin A) (\sec A + \tan A) \)

First, expand the expression:

\( = \sec A (1 – \sin A) \sec A + \sec A (1 – \sin A) \tan A \)

\( = \sec^2 A (1 – \sin A) + \sec A \tan A (1 – \sin A) \)

Factor out \( (1 – \sin A) \):

\( = (1 – \sin A)(\sec^2 A + \sec A \tan A) \)

Express in terms of cosine and sine:

\( = (1 – \sin A)\left( \frac{1}{\cos^2 A} + \frac{\sin A}{\cos^2 A} \right) = (1 – \sin A)\left( \frac{1 + \sin A}{\cos^2 A} \right) \)

Numerator becomes \( 1 – \sin^2 A = \cos^2 A \):

\( = \frac{\cos^2 A}{\cos^2 A} = 1 \)

7. Prove that \( (\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A \)

Expand both squares:

\( = \sin^2 A + 2\sin A \csc A + \csc^2 A + \cos^2 A + 2\cos A \sec A + \sec^2 A \)

Simplify using identities:

\( \sin A \csc A = 1 \), \( \cos A \sec A = 1 \)

\( \csc^2 A = 1 + \cot^2 A \), \( \sec^2 A = 1 + \tan^2 A \)

Also, \( \sin^2 A + \cos^2 A = 1 \)

Substitute:

\( = 1 + 2(1) + (1 + \cot^2 A) + 2(1) + (1 + \tan^2 A) \)

Combine like terms:

\( = 1 + 2 + 1 + \cot^2 A + 2 + 1 + \tan^2 A \)

\( = 7 + \tan^2 A + \cot^2 A \)

8. Simplify \( (1 – \cos \theta) (1 + \cos \theta) (1 + \cot^2 \theta) \)

First two terms are difference of squares:

\( = (1 – \cos^2 \theta)(1 + \cot^2 \theta) \)

\( = \sin^2 \theta \cdot \csc^2 \theta \) (since \( 1 + \cot^2 \theta = \csc^2 \theta \))

\( = \sin^2 \theta \cdot \frac{1}{\sin^2 \theta} = 1 \)

9. If \( \sec \theta + \tan \theta = p \), then what is the value of \( \sec \theta – \tan \theta \)?

We know the identity:

\( \sec^2 \theta – \tan^2 \theta = 1 \)

This can be written as:

\( (\sec \theta + \tan \theta)(\sec \theta – \tan \theta) = 1 \)

Given \( \sec \theta + \tan \theta = p \), then:

\( p (\sec \theta – \tan \theta) = 1 \)

Therefore:

\( \sec \theta – \tan \theta = \frac{1}{p} \)

10. If \( \csc \theta + \cot \theta = k \), then prove that \( \cos \theta = \frac{k^2 – 1}{k^2 + 1} \)

We know the identity:

\( \csc^2 \theta – \cot^2 \theta = 1 \)

This can be written as:

\( (\csc \theta + \cot \theta)(\csc \theta – \cot \theta) = 1 \)

Given \( \csc \theta + \cot \theta = k \), then \( \csc \theta – \cot \theta = \frac{1}{k} \)

Add the two equations:

\( 2\csc \theta = k + \frac{1}{k} = \frac{k^2 + 1}{k} \)

Subtract the two equations:

\( 2\cot \theta = k – \frac{1}{k} = \frac{k^2 – 1}{k} \)

Now, \( \cos \theta = \frac{\cot \theta}{\csc \theta} = \frac{\frac{k^2 – 1}{2k}}{\frac{k^2 + 1}{2k}} = \frac{k^2 – 1}{k^2 + 1} \)

10th Maths Trigonometry Exercise 11.3 Solutions

Exercise 11.3 Solutions

Exercise 11.3 Solutions

Class X Mathematics textbook by the State Council of Educational Research and Training, Telangana, Hyderabad

1. Evaluate:
(i) \(\frac{\tan 36^\circ}{\cot 54^\circ}\)

We know that \(\cot(90^\circ – \theta) = \tan\theta\), so \(\cot 54^\circ = \tan 36^\circ\).

Therefore, \(\frac{\tan 36^\circ}{\cot 54^\circ} = \frac{\tan 36^\circ}{\tan 36^\circ} = 1\)

(ii) \(\cos 12^\circ – \sin 78^\circ\)

We know that \(\sin(90^\circ – \theta) = \cos\theta\), so \(\sin 78^\circ = \cos 12^\circ\).

Therefore, \(\cos 12^\circ – \sin 78^\circ = \cos 12^\circ – \cos 12^\circ = 0\)

(iii) \(\csc 31^\circ – \sec 59^\circ\)

We know that \(\sec(90^\circ – \theta) = \csc\theta\), so \(\sec 59^\circ = \csc 31^\circ\).

Therefore, \(\csc 31^\circ – \sec 59^\circ = \csc 31^\circ – \csc 31^\circ = 0\)

(iv) \(\sin 15^\circ \sec 75^\circ\)

We know that \(\sec\theta = \frac{1}{\cos\theta}\) and \(\cos(90^\circ – \theta) = \sin\theta\).

\(\sec 75^\circ = \frac{1}{\cos 75^\circ} = \frac{1}{\sin 15^\circ}\)

Therefore, \(\sin 15^\circ \sec 75^\circ = \sin 15^\circ \times \frac{1}{\sin 15^\circ} = 1\)

(vi) \(\tan 26^\circ \tan 64^\circ\)

We know that \(\tan(90^\circ – \theta) = \cot\theta\), so \(\tan 64^\circ = \cot 26^\circ\).

Also, \(\cot\theta = \frac{1}{\tan\theta}\), so \(\tan 26^\circ \tan 64^\circ = \tan 26^\circ \cot 26^\circ = \tan 26^\circ \times \frac{1}{\tan 26^\circ} = 1\)

2. Show that:
(i) \(\tan 48^\circ \tan 16^\circ \tan 42^\circ \tan 74^\circ = 1\)

We can rearrange the terms:

\(\tan 48^\circ \tan 42^\circ \tan 16^\circ \tan 74^\circ\)

We know that \(\tan(90^\circ – \theta) = \cot\theta\), so:

\(\tan 42^\circ = \cot 48^\circ\) and \(\tan 74^\circ = \cot 16^\circ\)

Now the expression becomes:

\(\tan 48^\circ \cot 48^\circ \tan 16^\circ \cot 16^\circ\)

Since \(\tan\theta \cot\theta = 1\), the expression simplifies to \(1 \times 1 = 1\)

(ii) \(\cos 36^\circ \cos 54^\circ – \sin 36^\circ \sin 54^\circ = 0\)

We know that \(\cos(90^\circ – \theta) = \sin\theta\), so \(\cos 54^\circ = \sin 36^\circ\).

Similarly, \(\sin 54^\circ = \cos 36^\circ\).

Substituting these values:

\(\cos 36^\circ \sin 36^\circ – \sin 36^\circ \cos 36^\circ = 0\)

3. If \(\tan 2A = \cot(A – 18^\circ)\), where \(2A\) is an acute angle. Find the value of \(A\).

We know that \(\cot(90^\circ – \theta) = \tan\theta\), so we can write:

\(\tan 2A = \cot(A – 18^\circ) = \tan(90^\circ – (A – 18^\circ)) = \tan(108^\circ – A)\)

Since \(2A\) is acute, both sides are equal:

\(2A = 108^\circ – A\)

\(3A = 108^\circ\)

\(A = 36^\circ\)

4. If \(\tan A = \cot B\) where \(A\) and \(B\) are acute angles, prove that \(A + B = 90^\circ\).

We know that \(\cot B = \tan(90^\circ – B)\), so:

\(\tan A = \tan(90^\circ – B)\)

Since both \(A\) and \(B\) are acute angles, we can equate the angles:

\(A = 90^\circ – B\)

Therefore, \(A + B = 90^\circ\)

5. If \(A\), \(B\) and \(C\) are interior angles of a triangle \(ABC\), then show that \(\tan\left(\frac{A + B}{2}\right) = \cot\frac{C}{2}\)

In any triangle, the sum of interior angles is \(180^\circ\):

\(A + B + C = 180^\circ\)

Therefore, \(A + B = 180^\circ – C\)

Divide both sides by 2:

\(\frac{A + B}{2} = 90^\circ – \frac{C}{2}\)

Now take the tangent of both sides:

\(\tan\left(\frac{A + B}{2}\right) = \tan\left(90^\circ – \frac{C}{2}\right) = \cot\frac{C}{2}\)

This completes the proof.

6. Express \(\sin 75^\circ + \cos 65^\circ\) in terms of trigonometric ratios of angles between \(0^\circ\) and \(45^\circ\).

We can express both terms using complementary angle identities:

\(\sin 75^\circ = \cos 15^\circ\) (since \(\sin\theta = \cos(90^\circ – \theta)\))

\(\cos 65^\circ = \sin 25^\circ\) (since \(\cos\theta = \sin(90^\circ – \theta)\))

Therefore, \(\sin 75^\circ + \cos 65^\circ = \cos 15^\circ + \sin 25^\circ\)

Note: Both 15° and 25° are between 0° and 45° as required.

10th Maths Trigonometry Exercise 11.2 Solutions

Exercise 11.2 Solutions – Class X Mathematics

Exercise 11.2 Solutions

Class X Mathematics textbook by the State Council of Educational Research and Training, Telangana, Hyderabad

1. Evaluate the following:

(i) \(\sin 45^\circ + \cos 45^\circ\)

Solution:

We know that \(\sin 45^\circ = \frac{1}{\sqrt{2}}\) and \(\cos 45^\circ = \frac{1}{\sqrt{2}}\)

\(\sin 45^\circ + \cos 45^\circ = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}\)

(ii) \(\frac{\sin 30^\circ + \tan 45^\circ – \csc 60^\circ}{\cot 45^\circ + \cos 60^\circ – \sec 30^\circ}\)

Solution:

First, let’s evaluate each trigonometric function:

  • \(\sin 30^\circ = \frac{1}{2}\)
  • \(\tan 45^\circ = 1\)
  • \(\csc 60^\circ = \frac{2}{\sqrt{3}}\)
  • \(\cot 45^\circ = 1\)
  • \(\cos 60^\circ = \frac{1}{2}\)
  • \(\sec 30^\circ = \frac{2}{\sqrt{3}}\)

Numerator: \(\frac{1}{2} + 1 – \frac{2}{\sqrt{3}} = \frac{3}{2} – \frac{2}{\sqrt{3}}\)

Denominator: \(1 + \frac{1}{2} – \frac{2}{\sqrt{3}} = \frac{3}{2} – \frac{2}{\sqrt{3}}\)

Thus, the expression becomes \(\frac{\frac{3}{2} – \frac{2}{\sqrt{3}}}{\frac{3}{2} – \frac{2}{\sqrt{3}}} = 1\)

(v) \(\frac{\sec^2 60^\circ – \tan^2 60^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}\)

Solution:

We know the trigonometric identities:

  • \(\sec^2 \theta – \tan^2 \theta = 1\)
  • \(\sin^2 \theta + \cos^2 \theta = 1\)

Numerator: \(\sec^2 60^\circ – \tan^2 60^\circ = 1\) (by identity)

Denominator: \(\sin^2 30^\circ + \cos^2 30^\circ = 1\) (by identity)

Thus, the expression becomes \(\frac{1}{1} = 1\)

2. Choose the right option and justify your choice:

(i) \(\frac{2 \tan 30^\circ}{1 + \tan^2 45^\circ} =\)

Options:

(a) \(\sin 60^\circ\) (b) \(\cos 60^\circ\) (c) \(\tan 30^\circ\) (d) \(\sin 30^\circ\)

Solution:

Evaluate each part:

\(\tan 30^\circ = \frac{1}{\sqrt{3}}\), \(\tan 45^\circ = 1\)

Numerator: \(2 \times \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}}\)

Denominator: \(1 + 1^2 = 2\)

Expression: \(\frac{2/\sqrt{3}}{2} = \frac{1}{\sqrt{3}} \approx 0.577\)

Now evaluate options:

(d) \(\sin 30^\circ = 0.5\)

But \(\frac{1}{\sqrt{3}} \approx 0.577\) which is \(\sin 30^\circ\) plus some difference

Actually, \(\frac{1}{\sqrt{3}} = \tan 30^\circ\), but that’s option (c)

However, \(\sin 30^\circ = 0.5\) and our result is 0.577, which matches none exactly

But \(\frac{1}{\sqrt{3}} = \tan 30^\circ\), so correct answer is (c) \(\tan 30^\circ\)

(ii) \(\frac{1 – \tan^2 45^\circ}{1 + \tan^2 45^\circ} =\)

Options:

(a) \(\tan 90^\circ\) (b) 1 (c) \(\sin 45^\circ\) (d) 0

Solution:

\(\tan 45^\circ = 1\)

Numerator: \(1 – 1^2 = 0\)

Denominator: \(1 + 1^2 = 2\)

Expression: \(\frac{0}{2} = 0\)

Correct answer is (d) 0

(iii) \(\frac{2 \tan 30^\circ}{1 – \tan^2 30^\circ} =\)

Options:

(a) \(\cos 60^\circ\) (b) \(\sin 60^\circ\) (c) \(\tan 60^\circ\) (d) \(\sin 30^\circ\)

Solution:

This is the double angle formula for tangent:

\(\frac{2 \tan \theta}{1 – \tan^2 \theta} = \tan 2\theta\)

Here \(\theta = 30^\circ\), so expression equals \(\tan 60^\circ\)

Correct answer is (c) \(\tan 60^\circ\)

3. Evaluate \(\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ\). What is the value of \(\sin(60^\circ + 30^\circ)\). What can you conclude?

Solution:

First part:

\(\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2} = \frac{3}{4} + \frac{1}{4} = 1\)

Second part:

\(\sin(60^\circ + 30^\circ) = \sin 90^\circ = 1\)

Conclusion:

This demonstrates the angle addition formula: \(\sin(A + B) = \sin A \cos B + \cos A \sin B\)

4. Is it right to say that \(\cos(60^\circ + 30^\circ) = \cos 60^\circ \cos 30^\circ – \sin 60^\circ \sin 30^\circ\)?

Solution:

Left side: \(\cos(60^\circ + 30^\circ) = \cos 90^\circ = 0\)

Right side: \(\cos 60^\circ \cos 30^\circ – \sin 60^\circ \sin 30^\circ = \frac{1}{2} \times \frac{\sqrt{3}}{2} – \frac{\sqrt{3}}{2} \times \frac{1}{2} = \frac{\sqrt{3}}{4} – \frac{\sqrt{3}}{4} = 0\)

Both sides equal 0, so the statement is correct. This demonstrates the cosine addition formula: \(\cos(A + B) = \cos A \cos B – \sin A \sin B\)

5. In right angle triangle \(\triangle PQR\), right angle is at \(Q, PQ = 6 \text{cm}\) and \(\angle RPQ = 60^\circ\). Determine the lengths of \(QR\) and \(PR\).

Diagram description:

Right triangle PQR with right angle at Q. Points are ordered P-Q-R clockwise or counterclockwise. PQ is the side adjacent to the 60° angle at P, QR is opposite the 60° angle, and PR is the hypotenuse.

Solution:

Given:

  • Right angle at Q
  • PQ = 6 cm
  • \(\angle RPQ = 60^\circ\)

Using trigonometric ratios:

\(\tan 60^\circ = \frac{QR}{PQ} \Rightarrow QR = PQ \times \tan 60^\circ = 6 \times \sqrt{3} = 6\sqrt{3} \text{ cm}\)

\(\cos 60^\circ = \frac{PQ}{PR} \Rightarrow PR = \frac{PQ}{\cos 60^\circ} = \frac{6}{0.5} = 12 \text{ cm}\)

Thus, QR = \(6\sqrt{3}\) cm and PR = 12 cm

6. In \(\triangle XYZ\), right angle is at \(Y, YZ = x\), and \(XZ = 2x\). Then determine \(\angle YXZ\) and \(\angle YZX\).

Diagram description:

Right triangle XYZ with right angle at Y. Points are ordered X-Y-Z clockwise or counterclockwise. YZ is one leg (length x), XY is the other leg, and XZ is the hypotenuse (length 2x).

Solution:

Given:

  • Right angle at Y
  • YZ = x
  • XZ = 2x (hypotenuse)

First, find XY using Pythagorean theorem:

\(XY = \sqrt{XZ^2 – YZ^2} = \sqrt{(2x)^2 – x^2} = \sqrt{4x^2 – x^2} = \sqrt{3x^2} = x\sqrt{3}\)

Now find angles:

\(\sin \angle YXZ = \frac{YZ}{XZ} = \frac{x}{2x} = \frac{1}{2} \Rightarrow \angle YXZ = 30^\circ\)

\(\angle YZX = 90^\circ – \angle YXZ = 90^\circ – 30^\circ = 60^\circ\)

Thus, \(\angle YXZ = 30^\circ\) and \(\angle YZX = 60^\circ\)

7. Is it right to say that \(\sin (A + B) = \sin A + \sin B\)? Justify your answer.

Solution:

No, this is not correct in general. The correct formula is \(\sin(A + B) = \sin A \cos B + \cos A \sin B\).

Counterexample: Let A = 30° and B = 60°

\(\sin(30^\circ + 60^\circ) = \sin 90^\circ = 1\)

But \(\sin 30^\circ + \sin 60^\circ = 0.5 + \frac{\sqrt{3}}{2} \approx 0.5 + 0.866 = 1.366\)

Clearly, 1 ≠ 1.366, so the statement is false.

The correct relationship is the angle addition formula: \(\sin(A + B) = \sin A \cos B + \cos A \sin B\)

10th Maths Trigonometry Exercise 11.1 Solutions

Exercise 11.1 Solutions

Exercise 11.1 Solutions

1. In right angle triangle ABC, 8 cm, 15 cm and 17 cm are the lengths of AB, BC and CA respectively. Then, find sin A, cos A and tan A.

Diagram description: Right-angled triangle ABC with right angle at B. AB is the side adjacent to angle A (8 cm), BC is the side opposite to angle A (15 cm), and CA is the hypotenuse (17 cm).

Given: AB = 8 cm (adjacent to ∠A), BC = 15 cm (opposite to ∠A), CA = 17 cm (hypotenuse)

\[ \sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{CA} = \frac{15}{17} \]

\[ \cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AB}{CA} = \frac{8}{17} \]

\[ \tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{BC}{AB} = \frac{15}{8} \]

2. The sides of a right angle triangle PQR are \( PQ = 7 \, \text{cm} \), \( PR = 25 \, \text{cm} \) and \( \angle Q = 90^\circ \) respectively. Then find, tan P – tan R.

Diagram description: Right-angled triangle PQR with right angle at Q. PQ is one leg (7 cm), QR is the other leg (to be calculated), and PR is the hypotenuse (25 cm).

First, find QR using Pythagoras theorem:

\[ QR = \sqrt{PR^2 – PQ^2} = \sqrt{25^2 – 7^2} = \sqrt{625 – 49} = \sqrt{576} = 24 \, \text{cm} \]

Now calculate trigonometric ratios:

\[ \tan P = \frac{\text{opposite}}{\text{adjacent}} = \frac{QR}{PQ} = \frac{24}{7} \]

\[ \tan R = \frac{\text{opposite}}{\text{adjacent}} = \frac{PQ}{QR} = \frac{7}{24} \]

\[ \tan P – \tan R = \frac{24}{7} – \frac{7}{24} = \frac{576 – 49}{168} = \frac{527}{168} \]

3. In a right angle triangle ABC with right angle at B, in which \( a = 24 \, \text{units}, b = 25 \, \text{units} \) and \( \angle BAC = \theta \). Then, find \( \cos \theta \) and \( \tan \theta \).

Diagram description: Right-angled triangle ABC with right angle at B. BC = 24 units (opposite to θ), AB is the adjacent side (to be calculated), and AC = 25 units (hypotenuse).

Given: BC = a = 24 units (opposite to θ), AC = b = 25 units (hypotenuse)

First, find AB using Pythagoras theorem:

\[ AB = \sqrt{AC^2 – BC^2} = \sqrt{25^2 – 24^2} = \sqrt{625 – 576} = \sqrt{49} = 7 \, \text{units} \]

Now calculate trigonometric ratios:

\[ \cos \theta = \frac{AB}{AC} = \frac{7}{25} \]

\[ \tan \theta = \frac{BC}{AB} = \frac{24}{7} \]

4. If \( \cos A = \frac{12}{13} \), then find \( \sin A \) and \( \tan A \, (A < 90^\circ) \).

Given: \( \cos A = \frac{12}{13} = \frac{\text{adjacent}}{\text{hypotenuse}} \)

Let adjacent side = 12k, hypotenuse = 13k

Find opposite side using Pythagoras theorem:

\[ \text{Opposite} = \sqrt{(13k)^2 – (12k)^2} = \sqrt{169k^2 – 144k^2} = \sqrt{25k^2} = 5k \]

Now calculate:

\[ \sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5k}{13k} = \frac{5}{13} \]

\[ \tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{5k}{12k} = \frac{5}{12} \]

5. If 3 \(\tan A = 4\), then find \( \sin A \) and \( \cos A \).

Given: \( 3 \tan A = 4 \) ⇒ \( \tan A = \frac{4}{3} = \frac{\text{opposite}}{\text{adjacent}} \)

Let opposite side = 4k, adjacent side = 3k

Find hypotenuse using Pythagoras theorem:

\[ \text{Hypotenuse} = \sqrt{(4k)^2 + (3k)^2} = \sqrt{16k^2 + 9k^2} = \sqrt{25k^2} = 5k \]

Now calculate:

\[ \sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4k}{5k} = \frac{4}{5} \]

\[ \cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{3k}{5k} = \frac{3}{5} \]

6. In \( \triangle ABC \) and \( \triangle XYZ \), if \( \angle A \) and \( \angle X \) are acute angles such that \( \cos A = \cos X \) then show that \( \angle A = \angle X \).

Let’s consider two right triangles ABC and XYZ where \( \angle B = 90^\circ \) and \( \angle Y = 90^\circ \).

For \( \triangle ABC \):

\[ \cos A = \frac{AB}{AC} \]

For \( \triangle XYZ \):

\[ \cos X = \frac{XY}{XZ} \]

Given \( \cos A = \cos X \), so \( \frac{AB}{AC} = \frac{XY}{XZ} = k \) (say)

Let \( AB = k \cdot AC \) and \( XY = k \cdot XZ \)

Using Pythagoras theorem in both triangles:

In \( \triangle ABC \): \( BC = \sqrt{AC^2 – AB^2} = \sqrt{AC^2 – k^2 AC^2} = AC \sqrt{1 – k^2} \)

In \( \triangle XYZ \): \( YZ = \sqrt{XZ^2 – XY^2} = \sqrt{XZ^2 – k^2 XZ^2} = XZ \sqrt{1 – k^2} \)

Now all corresponding sides are proportional:

\[ \frac{AB}{XY} = \frac{k AC}{k XZ} = \frac{AC}{XZ} \]

\[ \frac{BC}{YZ} = \frac{AC \sqrt{1 – k^2}}{XZ \sqrt{1 – k^2}} = \frac{AC}{XZ} \]

Thus, \( \triangle ABC \sim \triangle XYZ \) by SSS similarity, and therefore \( \angle A = \angle X \).

7. Given \( \cot \theta = \frac{7}{8} \), then evaluate
(i) \( \frac{(1 + \sin \theta)(1 – \sin \theta)}{(1 + \cos \theta)(1 – \cos \theta)} \)
(ii) \( \frac{(1 + \sin \theta)}{\cos \theta} \)

Given: \( \cot \theta = \frac{7}{8} = \frac{\text{adjacent}}{\text{opposite}} \)

Let adjacent side = 7k, opposite side = 8k

Find hypotenuse:

\[ \text{Hypotenuse} = \sqrt{(7k)^2 + (8k)^2} = \sqrt{49k^2 + 64k^2} = \sqrt{113k^2} = k\sqrt{113} \]

Thus:

\[ \sin \theta = \frac{8}{\sqrt{113}}, \quad \cos \theta = \frac{7}{\sqrt{113}} \]

(i) Simplify the expression:

\[ \frac{(1 + \sin \theta)(1 – \sin \theta)}{(1 + \cos \theta)(1 – \cos \theta)} = \frac{1 – \sin^2 \theta}{1 – \cos^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} = \cot^2 \theta = \left(\frac{7}{8}\right)^2 = \frac{49}{64} \]

(ii) Evaluate:

\[ \frac{1 + \sin \theta}{\cos \theta} = \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta} = \sec \theta + \tan \theta \]

\[ = \frac{\sqrt{113}}{7} + \frac{8}{7} = \frac{8 + \sqrt{113}}{7} \]

8. In a right angle triangle ABC, right angle is at B. If \( \tan A = \sqrt{3} \), then find the value of
(i) \( \sin A \, \cos C + \cos A \, \sin C \)
(ii) \( \cos A \, \cos C – \sin A \, \sin C \)

Diagram description: Right-angled triangle ABC with right angle at B. Angle A is θ, angle C is (90°-θ). AB is the adjacent side to angle A, BC is the opposite side to angle A, and AC is the hypotenuse.

Given: \( \tan A = \sqrt{3} = \frac{\text{opposite}}{\text{adjacent}} = \frac{BC}{AB} \)

Let AB = 1 unit, BC = √3 units

Find hypotenuse AC:

\[ AC = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2 \, \text{units} \]

Now find trigonometric ratios:

\[ \sin A = \frac{BC}{AC} = \frac{\sqrt{3}}{2}, \quad \cos A = \frac{AB}{AC} = \frac{1}{2} \]

Since \( \angle C = 90^\circ – \angle A \):

\[ \sin C = \cos A = \frac{1}{2}, \quad \cos C = \sin A = \frac{\sqrt{3}}{2} \]

(i) Evaluate:

\[ \sin A \cos C + \cos A \sin C = \left(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2} \times \frac{1}{2}\right) = \frac{3}{4} + \frac{1}{4} = 1 \]

(ii) Evaluate:

\[ \cos A \cos C – \sin A \sin C = \left(\frac{1}{2} \times \frac{\sqrt{3}}{2}\right) – \left(\frac{\sqrt{3}}{2} \times \frac{1}{2}\right) = \frac{\sqrt{3}}{4} – \frac{\sqrt{3}}{4} = 0 \]

10th Maths Mensuration Exercise 10.4 Solutions

Exercise 10.4 Solutions

Exercise 10.4 Solutions

Problem 1: A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Diagram Description: A sphere with radius 4.2 cm is transformed into a cylinder with radius 6 cm. The volume remains the same during this transformation.

Solution:

Volume of sphere = Volume of cylinder

\(\frac{4}{3}\pi r^3 = \pi R^2 h\)

\(\frac{4}{3}\pi (4.2)^3 = \pi (6)^2 h\)

\(\frac{4}{3} \times 74.088 = 36 h\)

\(98.784 = 36 h\)

\(h = \frac{98.784}{36} = 2.744 \text{ cm}\)

Answer: The height of the cylinder is 2.744 cm.

Problem 2: Three metallic spheres of radii 6 cm, 8 cm and 10 cm respectively are melted together to form a single solid sphere. Find the radius of the resulting sphere.
Diagram Description: Three separate spheres with radii 6 cm, 8 cm, and 10 cm are combined to form one larger sphere. The total volume of the three spheres equals the volume of the new sphere.

Solution:

Total volume = Volume of sphere 1 + Volume of sphere 2 + Volume of sphere 3

\(\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (6)^3 + \frac{4}{3}\pi (8)^3 + \frac{4}{3}\pi (10)^3\)

\(r^3 = 6^3 + 8^3 + 10^3 = 216 + 512 + 1000 = 1728\)

\(r = \sqrt[3]{1728} = 12 \text{ cm}\)

Answer: The radius of the resulting sphere is 12 cm.

Problem 3: A 20 m deep well of diameter 7 m is dug and the earth got by digging is evenly spread out to form a rectangular platform of base 22 m × 14 m. Find the height of the platform.
Diagram Description: A cylindrical well with diameter 7 m and depth 20 m is dug. The excavated earth forms a rectangular platform with dimensions 22 m × 14 m × height h m.

Solution:

Volume of earth dug = Volume of well = \(\pi r^2 h = \pi (3.5)^2 \times 20\)

Volume of platform = \(22 \times 14 \times h\)

\(\pi (3.5)^2 \times 20 = 22 \times 14 \times h\)

\(\frac{22}{7} \times 12.25 \times 20 = 308 h\)

\(770 = 308 h\)

\(h = \frac{770}{308} = 2.5 \text{ m}\)

Answer: The height of the platform is 2.5 m.

Problem 4: A well of diameter 14 m is dug 15 m deep. The earth taken out of it has been spread evenly to form circular embankment all around the wall of width 7 m. Find the height of the embankment.
Diagram Description: A cylindrical well with diameter 14 m and depth 15 m is dug. The excavated earth forms a circular embankment (a ring-shaped structure) around the well with width 7 m and height h m.

Solution:

Volume of earth dug = \(\pi (7)^2 \times 15\)

Outer radius of embankment = 7 m (well radius) + 7 m (width) = 14 m

Volume of embankment = \(\pi (14^2 – 7^2) h = \pi (196 – 49) h = 147\pi h\)

\(\pi \times 49 \times 15 = 147\pi h\)

\(h = \frac{49 \times 15}{147} = 5 \text{ m}\)

Answer: The height of the embankment is 5 m.

Problem 5: A container shaped a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, making a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
Diagram Description: A cylindrical container with diameter 12 cm and height 15 cm contains ice cream. The ice cream is to be distributed into cones with height 12 cm and diameter 6 cm, each topped with a hemispherical scoop (radius 3 cm).

Solution:

Volume of ice cream in cylinder = \(\pi (6)^2 \times 15 = 540\pi \text{ cm}^3\)

Volume of one cone = \(\frac{1}{3}\pi (3)^2 \times 12 = 36\pi \text{ cm}^3\)

Volume of hemisphere = \(\frac{2}{3}\pi (3)^3 = 18\pi \text{ cm}^3\)

Total volume per cone = \(36\pi + 18\pi = 54\pi \text{ cm}^3\)

Number of cones = \(\frac{540\pi}{54\pi} = 10\)

Answer: 10 cones can be filled with the ice cream.

Problem 6: How many silver coins, 1.75 cm in diameter and thickness 2 mm, need to be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
Diagram Description: Multiple cylindrical coins (each with diameter 1.75 cm and thickness 2 mm) are melted to form a rectangular cuboid with dimensions 5.5 cm × 10 cm × 3.5 cm.

Solution:

Volume of one coin = \(\pi (0.875)^2 \times 0.2 \approx 0.481 \text{ cm}^3\)

Volume of cuboid = \(5.5 \times 10 \times 3.5 = 192.5 \text{ cm}^3\)

Number of coins = \(\frac{192.5}{0.481} \approx 400\)

Answer: Approximately 400 coins are needed.

Problem 7: A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top is 5 cm. It is filled with water up to the rim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, \(\frac{1}{4}\) of the water flows out. Find the number of lead shots dropped into the vessel.
Diagram Description: An inverted cone-shaped vessel with height 8 cm and top radius 5 cm is completely filled with water. Small spherical lead shots (each with radius 0.5 cm) are dropped into the vessel, displacing 1/4 of the water volume.

Solution:

Volume of cone = \(\frac{1}{3}\pi (5)^2 \times 8 = \frac{200}{3}\pi \text{ cm}^3\)

Volume of water displaced = \(\frac{1}{4} \times \frac{200}{3}\pi = \frac{50}{3}\pi \text{ cm}^3\)

Volume of one lead shot = \(\frac{4}{3}\pi (0.5)^3 = \frac{\pi}{6} \text{ cm}^3\)

Number of lead shots = \(\frac{\frac{50}{3}\pi}{\frac{\pi}{6}} = 100\)

Answer: 100 lead shots were dropped into the vessel.

Problem 8: A solid metallic sphere of diameter 28 cm is melted and recast into a number of smaller cones, each of diameter 4 \(\frac{2}{3}\) cm and height 3 cm. Find the number of cones so formed.
Diagram Description: A large metallic sphere with diameter 28 cm is melted and reshaped into multiple smaller cones. Each cone has diameter 14/3 cm (4 2/3 cm) and height 3 cm.

Solution:

Radius of sphere = 14 cm

Volume of sphere = \(\frac{4}{3}\pi (14)^3 = \frac{10976}{3}\pi \text{ cm}^3\)

Radius of each cone = \(\frac{14}{6} = \frac{7}{3} \text{ cm}\)

Volume of one cone = \(\frac{1}{3}\pi \left(\frac{7}{3}\right)^2 \times 3 = \frac{49}{9}\pi \text{ cm}^3\)

Number of cones = \(\frac{\frac{10976}{3}\pi}{\frac{49}{9}\pi} = \frac{10976}{3} \times \frac{9}{49} = 672\)

Answer: 672 cones can be formed.

10th Maths Mensuration Exercise 10.3 Solutions

10th Maths Mensuration Exercise 10.2 Solutions

Exercise 10.2 Solutions – Class X Mathematics

Exercise 10.2 Solutions

Class X Mathematics textbook by the State Council of Educational Research and Training, Telangana, Hyderabad

Problem 1

A toy is in the form of a cone mounted on a hemisphere of the same diameter. The diameter of the base and the height of the cone are 6 cm and 4 cm respectively. Determine the surface area of the toy. [use π = 3.14]

[Diagram description: Cone with height 4cm mounted on hemisphere with diameter 6cm]

Solution:

Given: Diameter = 6 cm ⇒ Radius (r) = 3 cm

Cone height (h) = 4 cm

Slant height of cone (l) = √(r² + h²) = √(9 + 16) = 5 cm

Curved surface area of cone = πrl = 3.14 × 3 × 5 = 47.1 cm²

Curved surface area of hemisphere = 2πr² = 2 × 3.14 × 9 = 56.52 cm²

Total surface area = Cone CSA + Hemisphere CSA = 47.1 + 56.52 = 103.62 cm²

Problem 2

A solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end. The radius of the common base is 8 cm and the heights of the cylindrical and conical portions are 10 cm and 6 cm respectively. Find the total surface area of the solid. [use π = 3.14]

[Diagram description: Cylinder (height 10cm) with hemisphere on one end and cone (height 6cm) on other end, all with radius 8cm]

Solution:

Given: Radius (r) = 8 cm

Cylinder height (h₁) = 10 cm, Cone height (h₂) = 6 cm

Cylinder CSA = 2πrh₁ = 2 × 3.14 × 8 × 10 = 502.4 cm²

Cone slant height (l) = √(r² + h₂²) = √(64 + 36) = 10 cm

Cone CSA = πrl = 3.14 × 8 × 10 = 251.2 cm²

Hemisphere CSA = 2πr² = 2 × 3.14 × 64 = 401.92 cm²

Total surface area = Cylinder CSA + Cone CSA + Hemisphere CSA = 502.4 + 251.2 + 401.92 = 1155.52 cm²

Problem 3

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the capsule is 14 mm and the thickness is 5 mm. Find its surface area.

[Diagram description: Cylinder with two hemispheres on both ends, total length 14mm, diameter 5mm]

Solution:

Given: Total length = 14 mm, Diameter = 5 mm ⇒ Radius (r) = 2.5 mm

Height of cylinder = Total length – 2 × radius = 14 – 5 = 9 mm

Cylinder CSA = 2πrh = 2 × 3.14 × 2.5 × 9 ≈ 141.3 mm²

Two hemispheres = 1 full sphere surface area = 4πr² = 4 × 3.14 × 6.25 ≈ 78.5 mm²

Total surface area = 141.3 + 78.5 ≈ 219.8 mm²

Problem 4

Two cubes each of volume 64 cm³ are joined end to end together. Find the surface area of the resulting cuboid.

Solution:

Volume of each cube = 64 cm³ ⇒ Side length (a) = ∛64 = 4 cm

When joined, cuboid dimensions become: Length = 8 cm, Breadth = 4 cm, Height = 4 cm

Total surface area = 2(lb + bh + hl) = 2(8×4 + 4×4 + 4×8) = 2(32 + 16 + 32) = 160 cm²

Problem 5

A storage tank consists of a circular cylinder with a hemisphere stuck on either end. If the external diameter of the cylinder be 1.4 m and its length be 8 m, find the cost of painting it on the outside at rate of ₹20 per m².

[Diagram description: Cylinder (length 8m) with two hemispheres on both ends, diameter 1.4m]

Solution:

Given: Diameter = 1.4 m ⇒ Radius (r) = 0.7 m, Cylinder height (h) = 8 m

Cylinder CSA = 2πrh = 2 × (22/7) × 0.7 × 8 = 35.2 m²

Two hemispheres = 1 full sphere surface area = 4πr² = 4 × (22/7) × 0.49 ≈ 6.16 m²

Total surface area = 35.2 + 6.16 = 41.36 m²

Cost of painting = 41.36 × 20 = ₹827.20

Problem 6

A sphere, a cylinder and a cone have the same radius and same height. Find the ratio of their volumes.

[Diagram description: Sphere, cylinder, and cone with same radius and height]

Solution:

Given: Same radius (r) and height (h), and for sphere: diameter = height ⇒ h = 2r

Volume of sphere = (4/3)πr³

Volume of cylinder = πr²h = πr²(2r) = 2πr³

Volume of cone = (1/3)πr²h = (1/3)πr²(2r) = (2/3)πr³

Ratio = Sphere : Cylinder : Cone = (4/3) : 2 : (2/3) = 4 : 6 : 2 = 2 : 3 : 1

Problem 7

A hemisphere is cut out from one face of a cubical wooden block such that the diameter of the hemisphere is equal to the side of the cube. Determine the total surface area of the remaining solid.

[Diagram description: Cube with hemisphere removed from one face, diameter of hemisphere equals side of cube]

Solution:

Let side of cube = a ⇒ Radius of hemisphere = a/2

Total surface area of cube = 6a²

Area removed (circle) = π(a/2)² = πa²/4

Curved surface area added by hemisphere = 2π(a/2)² = πa²/2

Net change = -πa²/4 + πa²/2 = +πa²/4

Total surface area = 6a² + πa²/4 = a²(6 + π/4)

Problem 8

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm and its radius of the base is 3.5 cm, find the total surface area of the article.

[Diagram description: Cylinder (height 10cm, radius 3.5cm) with hemispheres scooped out from both ends]

Solution:

Given: Radius (r) = 3.5 cm, Cylinder height = 10 cm

Cylinder CSA = 2πrh = 2 × (22/7) × 3.5 × 10 = 220 cm²

Two hemispheres = 1 full sphere surface area = 4πr² = 4 × (22/7) × 12.25 = 154 cm²

Area of two circular tops removed = 2 × πr² = 2 × (22/7) × 12.25 = 77 cm²

Net surface area = Cylinder CSA + Sphere CSA – Removed circular areas = 220 + 154 – 77 = 297 cm²