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 an angle \(\alpha\) at the centre of the park, the height of the pole is :

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

Hint:

Find the chord length AB from the central angle, then use the right triangle formed by the pole and the line of sight from B.

Answer 1

The Correct Option is A

Let O be the centre. In triangle OAB, \(OA=OB=a\) and \(\angle AOB=\alpha\), so the chord \(AB=2a\sin\frac{\alpha}{2}\).
The pole at A is vertical, so the triangle formed by the pole, AB and the line of sight from B is right-angled at A, with angle of elevation \(\alpha\) at B.
Height \(h=AB\tan\alpha=2a\sin\frac{\alpha}{2}\tan\alpha\).
Answer: (1) \(2 a \sin \frac{\alpha}{2} \tan \alpha\)