Question

A vertical pole stands at a point $A$ on the boundary of a circular park of radius $a$ and subtends an angle $\alpha$ at another point $B$ on the boundary. If the chord $AB$ subtends angle $\alpha$ at the centre of the park, then the height of the pole is

• A. $2a\cos \frac{\alpha }{2}\tan \alpha$
• B. $2a\sin \frac{\alpha }{2}\tan \alpha$
• C. $2a\cos \frac{\alpha }{2}\cot \alpha$
• D. $2a\sin \frac{\alpha }{2}\cot \alpha$

Answer 1

The correct option is B $2a\sin \frac{\alpha }{2}\tan \alpha$

Let $O$ be centre of the park and $h$ be the height of pole, i.e., $AC=h$


Since $OA=OB=a,$
$\therefore \angle OAB=\angle OBA=\beta$ (say)
$\beta +\beta +\alpha ={180}^{\circ }$
$\Rightarrow \beta ={90}^{\circ }-\frac{\alpha }{2}$

Using sine rule in $△OAB,$
$\frac{AB}{\sin \alpha }=\frac{OA}{\sin \beta }$
$\Rightarrow AB=a\times \frac{\sin \alpha }{\sin ({90}^{\circ }-\frac{\alpha }{2})}$
$\Rightarrow AB=2a\sin \frac{\alpha }{2}$
Now, in $△ABC,$
$\tan \alpha =\frac{AC}{AB}$
$\therefore h=2a\sin \frac{\alpha }{2}\tan \alpha$