Question

The angles of elevation of the top of a tower from two points at distance of 5 metres and 20 metres from the base of the tower and is the same straight line with it, are complementary. Find the height of the tower.

Answer 1

Let the height of the tower be AB.
We have.
AC = 5m, AD = 20m
Let the angle of elevation of the top of the tower (i.e. ∠ACB) from point C be θ .
Then,
the angle of elevation of the top of the tower (i.e. Z ADB) from point D
=(90° -θ)
Now, in ΔABC

`tan theta = (AB)/(AC)`

`⇒ tan theta = (AB)/5`            ...................(i)

Also, in ΔABD,

`cot (90° - theta ) = (AD)/(AB)`

`⇒  tan theta = 20/(AB)`                .................(ii)

From (i) and (ii), we get

`(AB)/5 = 20/(AB)`

`⇒ AB^2 = 100`

`⇒  AB = sqrt(100)`

∴ AB = 10 m

So, the height of the tower is 10 m.