Question

An observer 1.7m tall, is 20√3 m away from a tower. The angle of elevation from
the eye of the observer to the top of the tower is 300
. Find the height of the tower

Answer 1

Step-by-step explanation:

Let AB be the height of the tower and DE be the height of the observer.

Then,

In Δ ACD ,

AC/DC = tan 30°

⇒ x/√30 = tan 30° = 1/√3

⇒ x = 20 m

AB = 20 m + 1.7 m

= 21.7 m

So, the height of the tower is 21.7 m