JEE Main · 2007 · 29 Apr 2007 · Offline · Heights and Distances

Let $O$ be the foot of the tower at the centre and $r$ the radius of the park. In triangle $OAB$ the sides $OA=OB=r$ and the chord $AB=a$ subtend $60^{\circ }$ at $O$, so the cosine rule gives $a^2=2r^2-2r^2\cos 60^{\circ }=r^2$, hence $r=a$ and the triangle is equilateral. The top of the tower is seen from A at $30^{\circ }$, so $\tan 30^{\circ }=\frac{h}{r}=\frac{h}{a}$, giving a height $h=\frac{a}{\sqrt{3}}$.