Question

A TV tower stands vertically on a bank of a canal. The tower is watched
from a point on the other bank directly opposite to it. The angle of elevation of the top
of the tower is 58°. From another point 20 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°. Find the
height of the tower and the width of the canal. (tan 58° = 1.6003)​

Answer 1

• Height of the tower is 18.07 m
• width of the canal is 11.29 m

Step-by-step explanation:

Let,

• AB be the height of the TV tower, CD = 20 m
• BC be the width of the canal

In the right angled triangle ΔABC,

⟶ $\sf{tan\:58°}$ = $\sf\dfrac{AB}{BC}$

⟶ $\sf{1.6003}$ = $\sf\dfrac{AB}{BC}$ --------- (1)

In the right angled ΔABD,

⟶ $\sf{tan\:30°}$ = $\sf\dfrac{AB}{BD}$ = $\sf\dfrac{AB}{BC\:+\:CD}$

⟶ $\sf\dfrac{1}{\sqrt{3}}$ = $\sf\dfrac{AB}{BC\:+\:CD}$--------- (2)

Dividing (1) & (2) we get,

⟶ $\sf\dfrac{1.6003}{\dfrac{1}{\sqrt{3}}}$ = $\sf\dfrac{BC\:+\:20}{BC}$

⟶ BC = $\sf\dfrac{20}{1.7717}$

⟶ $\sf{BC\:=\:11.29\:m}$ -------- (3)

⟶ $\sf{1.6003\:=\dfrac{AB}{11.29}}$ (from (1) & (3))

⟶ $\sf{AB\:=\:18.07}$

• Hence, the height of the tower is 18.07 m and width of the canal is 11.29 m.