A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of `60^0` with the wall, then the height of the wall is `15sqrt(3)m` (b) `(15sqrt(3))/2m` (c) `(15)/2m` (d) `15 m`

Explanation

Given that, the height of the ladder = 15m
Let the height of the vertical wall = h
and the ladder makes an angle of elevation `60^(@)` with the wall i.e., `theta=60^(@)`.
In `DeltaQPR, cos60^(@)=(PR)/(PQ)=h/(15)`
`rArr 1/2=h/15`
`rArr h=15/2 m`.
Hence, the required height of the wall `15/2`m.