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