Trigonometrical Ratios — Case-Study Based Questions

Case Study:

One day three friends Amit(A), Binay(B) and Chanchal(C) were playing hide and seek game in the park of their society. Amit and Binay hide in the shrubs and Chanchal has to find both of them. If the position of three friends are at A, B and C respectively, as shown in the figure and forms a right angled triangle ABC such that AB = 6 m, BC = $2\sqrt{3}$ m and ∠B = 90°.

Based on the above information answer the following questions:

1. The length of AC is :
   (a) 4 m
   (b) $8\sqrt{3}$ m
   (c) $5\sqrt{3}$ m
   (d) $4\sqrt{3}$ m

2. The measure of ∠A is :
   (a) 30°
   (b) 45°
   (c) 60°
   (d) 90°

3. The measure of ∠C is :
   (a) 30°
   (b) 45°
   (c) 60°
   (d) 90°

4. cos 2A is equal to :
   (a) 1
   (b) $\dfrac{1}{2}$
   (c) $\dfrac{\sqrt{3}}{2}$
   (d) $\sqrt{3}$

5. $2\sin(\dfrac{C}{2})$ is equal to :
   (a) 1
   (b) $\dfrac{1}{2}$
   (c) $\sqrt{3}$
   (d) $\dfrac{\sqrt{3}}{2}$

Answer

1. Perpendicular = BC = $2\sqrt{3}$ m and Base = AB = 6 m

Hypotenuse = AC

We will find Hypotenuse by using Pythagoras Theorem,

Hypotenuse² = Perpendicular² + Base²

Hypotenuse² = $(2\sqrt{3})^2$ + 6²

Hypotenuse² = 12 + 36

Hypotenuse² = 48

Hypotenuse = $\sqrt{48} = 4\sqrt{3}$ m

Hence, option (d) is the correct option.

2. sin A = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

sin A = $\dfrac{BC}{AC} = \dfrac{2\sqrt{3}}{4\sqrt{3}}$

sin A = $\dfrac{1}{2}$

sin A = sin 30°

A = 30°

Hence, option (a) is the correct option.

3. sin C = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

sin C = $\dfrac{AB}{AC} = \dfrac{6}{4\sqrt{3}}$

sin C = $\dfrac{\sqrt{3}}{2}$

sin C = sin 60°

C = 60°

Hence, option (c) is the correct option.

4. As, A = 30°

cos 2(30°) = cos 60° = $\dfrac{1}{2}$

Hence, option (b) is the correct option.

5. As, C = 60°

$2\sin(\dfrac{C}{2}) = 2\sin(\dfrac{60}{2})$

= 2 sin 30°

= $2 \times \dfrac{1}{2}$

= 1.

Hence, option (a) is the correct option.

Case Study:

In the figure given below, the rod AB of length 4 inches of a TV disc antena is fixed at right angle to the wall and a rod BC of length 8 inches is supporting the disc.

Based on the above information answer the following questions:

1. The measure of ∠ACB is :
   (a) 30°
   (b) 45°
   (c) 60°
   (d) 90°

2. The value of tan ∠ABC is :
   (a) $\dfrac{1}{\sqrt{3}}$
   (b) $\sqrt{3}$
   (c) 1
   (d) 0

3. The value of sin²∠ACB + sin²∠ABC is :
   (a) $\dfrac{1}{2}$
   (b) 0
   (c) 1
   (d) not defined

4. The length of AC is :
   (a) 6 inches
   (b) $5\sqrt{3}$ inches
   (c) $4\sqrt{2}$ inches
   (d) $4\sqrt{3}$ inches

5. The value of sin ∠ACB + cos ∠ABC + cot ∠BAC is :
   (a) 0
   (b) 1
   (c) 2
   (d) not defined

Answer

1. AB = Perpendicular = 4 inches

BC = Hypotenuse = 8 inches

sin∠ACB = $\dfrac{\text Perpendicular}{\text Hypotenuse}$

sin∠ACB = $\dfrac{4}{8} = \dfrac{1}{2}$

sin∠ACB = sin 30°

∠ACB = 30°

Hence, option (a) is the correct option.

2. ∠ABC + ∠BAC + ∠ACB = 180°

∠ABC + 90° + 30° = 180°

∠ABC = 180° - 120° = 60°

tan ∠ABC = tan 60° = $\sqrt{3}$.

Hence, option (b) is the correct option.

3. sin²∠ACB + sin²∠ABC

= sin²30° + sin²60°

= $\Big(\dfrac{1}{2}\Big)^2 + \Big(\dfrac{\sqrt{3}}{2}\Big)^2$

= $\dfrac{1}{4} + \dfrac{3}{4}$

= $\dfrac{4}{4}$

= 1.

Hence, option (c) is the correct option.

4. AC = Base

By pythagoras theorem,

Hypotenuse² = Base² + Perpendicular²

Base² = Hypotenuse² - Perpendicular²

AC² = BC² - AB²

AC² = 8² - 4²

AC² = 64 - 16

AC² = 48

AC = $\sqrt{48}$

AC = $4\sqrt{3}$ inches.

Hence, option (d) is the correct option.

5. Solving,

sin ∠ACB + cos ∠ABC + cot ∠BAC

= sin 30° + cos 60° + cot 90°

= $\dfrac{1}{2} + \dfrac{1}{2} + 0$

= 1.

Hence, option (b) is the correct option.