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`

Explanation

`BC=x`

`AB=y`

`tan60^@=AB/BC=y/x`

`sqrt3=y/x`

`x=y/sqrt3`

`tan45^@=(AB)/(BD)=(AB)/(DC+CB)`

`=y/(7+x)`

`=1`

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