Question

A tower is situated on horizontal plane. From two points, the line joining these points passes through the base and which are a andb distance from the base. The angle of elevation of the top are $90 ^ { \circ } - \alpha$and$\theta$ is that angle which two

points joining the line makes at the top, the height of tower

will be

• A. $\frac { a + b } { a - b }$
• B. b $\frac { a - b } { a + b }$
• C. $\sqrt { a b }$
• D. $( a b ) ^ { \frac { 1 } { 3 } }$

Answer 1

Answer: (C)

$\sqrt { a b }$

Answer 2

Let there are two points C and D on horizontal line passing from point B of the base of the tower AB. The distance of these points are b and a from B respectively i.e., $B C = b$

$\therefore$Line CD, on the top of tower A subtends an angle $\theta$, hence $\alpha$ and $90 ^ { \circ } - \alpha$.

$\angle B D A = 90 ^ { \circ } - \alpha$

In $\triangle A B C , A B = B C \tan \alpha = b \tan \alpha$ .........(i)

And in $\triangle A B D , A B = B D \tan \left( 90 ^ { \circ } - \alpha \right) = a \cot \alpha$ .........(ii)

Multiplying equation (i) and (ii) $( A B ) ^ { 2 } = ( b \tan \alpha ) ( a \cot \alpha ) = a b , \Rightarrow A B = \sqrt { ( a b ) }$