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?

Diagram Description: A right triangle where the vertical side represents the tower (height h), the horizontal side represents the ground distance (15 m), and the angle between the ground and the line of sight to the top of the tower is 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:

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

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

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

Answer: 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.

Diagram Description: A tree broken into two parts. The lower part is vertical (height h), the upper part forms a 30° angle with the ground and touches the ground 6m from the base. The total height before breaking would be h + the length of the broken part.

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

Answer: 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?

Diagram Description: A right triangle where the vertical side is 2m (height of slide), the hypotenuse is the slide (length L), and the angle between the ground and slide is 30°.

Solution:

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

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

Using trigonometric ratio:

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

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

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

Answer: 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?

Diagram Description: A vertical pole (15m) casting a shadow (15√3m) on the ground, forming a right triangle. The angle of elevation of the sun is the angle between the ground and the line from the top of the shadow to the top of the pole.

Solution:

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

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

Using trigonometric ratio:

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

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

\[ \theta = 30^\circ \]

Answer: 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?

Diagram Description: A vertical pole (10m) with three ropes extending from the top to the ground, each making a 30° angle with the pole. The length of each rope is the hypotenuse of a right triangle where the pole is one side.

Solution:

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

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

Using trigonometric ratio:

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

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

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

Answer: 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?

Diagram Description: A building of height 6m with a person at the top. A line of sight to a target on the ground makes a 60° angle of depression (which equals the angle of elevation from the target to the person).

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:

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

Answer: 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?

Diagram Description: A pole (9m) with a ladder leaning against it. The ladder reaches 1.8m below the top (so 7.2m up the pole). The ladder makes a 60° angle with the ground, forming a right triangle with the pole and 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:

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

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

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

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

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

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

Answer: 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?

Diagram Description: A river with two parallel banks. The boat’s path forms a 60° angle with the near bank, traveling 600m diagonally. The width of the river is the perpendicular distance between the banks.

Solution:

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

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

Using trigonometric ratio:

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

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

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

Answer: 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?

Diagram Description: An observer (1.8m tall) standing 13.2m from a palm tree. The line of sight from the observer’s eyes to the top of the tree makes a 45° angle with the horizontal. The total tree height is the observer’s height plus the height above the observer’s eyes.

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

Answer: 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.

Diagram Description: Triangle ABC with sides AB = 5cm, AC = 6cm, and angle between them (∠BAC) = 30°. The area can be found using the formula involving two sides and the included angle.

Solution:

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

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

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

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

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

Answer: The area of the triangle is 7.5 cm².

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