AB is a vertical pole with B at the ground level and A at the top. A man finds that the angle of elevation of the point A from a certain point C on the ground is `60o` . He moves away from the pole along the line BC to a point D such that `C D""=""7""m` . From D the angle of elevation of the point A is `45o` . Then the height of the pole is (1) `(7sqrt(3))/2dot1/(sqrt(3)-1)m` (2) `(7sqrt(3))/2dot(sqrt(3)+1)m` (3) `(7sqrt(3))/2dot(sqrt(3)-1)m` (4) `(7sqrt(3))/2dot1/(sqrt(3)+1)m`

Options

A. `(7sqrt3)/2(sqrt3-1)m`
B. `(7sqrt3)/2(1/(sqrt3+1))` m
C. `(7sqrt3)/2 (1/(sqrt3-1))m`
D. `(7sqrt3)/2(sqrt3+1)` m

Answer

D

Explanation

Let h be the height of pole AB.

In `triangle's` ABC and ABD, we have
`tan 60^@=h/(BC)` and `tan 45^@=h/(BC+7)`
`BC=h/sqrt3` and `BC+7=h`
`rArr h/sqrt3+7=h rArr h ((sqrt3-1)/sqrt3)=7 rArr h = (7sqrt3)/2 (sqrt3+1)` m