An observer, `1.5 m` tall, is `20.5 m` away from a tower `22 m` high. Determine the angle of elevation of the top of the tower from the eye of the observer.

Options

A. `60^@`
B. `45^@`
C. `30^@`
D. `90^@`

Answer

B

Explanation

Let the anlge of elevation of the top of the tower from the eye of the observer is `theta`.
Given that, AB=22m, PQ=1.5m =MB
and QB=PM=20.5 m
`rArr AM=AB-MB`
`=22-1.5=20.5`m
Now, in `DeltaAPM, tan theta=(AM)/(PM) = (20.5)/(20.5)=1`
`rArr tantheta= tan45^(@)`
`therefore theta=45^(@)`
Hence, required angle of elevation of the top of the tower from the eye of the observer is `45^(@)`