Radio towers are used for transmitting a range of communication

Leave a Comment / Class 10th, Class 10th Case Study, trigonometry applications / By Sapling Academy

Q) Radio towers are used for transmitting a range of communication services including radio and television. The tower will either act as an antenna itself or support one or more antennas on its structure.

On a similar concept, a radio station tower was built in two stations A and B (B vertically below A). The tower is supported by wires AO and BO from a point O on the ground. Distance between the base C of the tower and the point O is 36 m. From O, the angles of elevation of the tops of station B and station A are 30º and 45º respectively.

Based on the above, answer the following questions :
(i) Find the height of station B.
(ii) Find the height of station A.
(iii) (a) Find the length of the wire OA.
OR
(iii) (b) Find the length of the wire OB.

Ans:

STEP BY STEP SOLUTION

First, we make a diagram for the question.

(i) Height of station B:

Let’s start from Δ BOC:

tan ∠ BOC = $\frac{BC}{OC}$

∴ $\tan 30 = \frac {X}{36}$

∴ $\frac{1}{\sqrt 3} = \frac {X}{36}$

∴ $X = \frac{36}{\sqrt 3}$

∴ X = 12 √3

Therefore, the height of tower B is 12 √3 m.

(ii) Height of station A:

In Δ AOC:

tan ∠ AOC = $\frac{AC}{OC}$

∴ $\tan 45 = \frac {H}{36}$

∴ 1 = $\frac {H}{36}$

∴ H = 36

Therefore, the height of tower A is 36 m.

(iii) (a) Length of wire OA:

In Δ AOC:

cos ∠ AOC = $\frac{OC}{OA}$

∴ $\cos 45 = \frac {36}{OA}$

∴ $\frac{1}{\sqrt2} = \frac{36}{OA}$

∴ OA = 36 √2

Therefore, the length of wire OA is 36 √2 m.

(iii) (b) Length of wire OB:

In Δ BOC:

cos ∠ BOC = $\frac{OC}{OB}$

∴ $\cos 30 = \frac {36}{OB}$

∴ $\frac{\sqrt3}{2} = \frac {36}{OB}$

∴ OB = $\frac{72}{\sqrt 3}$

∴ OB = 24 √3

Therefore, the length of wire OA is 24 √3 m.

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