Question

A tower subtends an angle $\alpha$ at point 'A' in the plane of its base and the angle of depression of the foot of the tower at height 'b' just above A is $\beta$. The height of the tower is

• A. $b\tan \alpha \cot \beta$
• B. $b\cot \alpha \cot \beta$
• C. $b\tan \alpha \tan \beta$
• D. $b\cot \alpha \tan \beta$

Answer 1

The correct option is A $b\tan \alpha \cot \beta$

Let $AP=BR=x$ then from $\Delta APQ$ we find that
the height of the tower $=PQ=x\tan \alpha =\left(b\cot \beta \right)\tan \alpha$
($\because$ In triangle APB, $\frac{x}{b}=\cot \beta$)