QJEEM07Ma79fc: JEE Main 2007 Mathematics - Class 12 JEEM07Ma79fc Solution: A tower stands at the centre of a circular park. and are two…

In the $\Delta AOB,\,\,\angle AOB = {60^ \circ },$ and

$\angle OBA = \angle OAB$

(since $OA=OB=AB$ radius of same circle).

$\therefore$ $\Delta AOB$ is a equilateral triangle.

Let the height of tower is $h$



Given distance between two points $A$ & $B$ lie on boundary of

circular park, subtends an angle of ${60^ \circ }$ at the foot of the tower

is $AB$ i.e. $AB$ $=a.$ A tower $OC$ stands at the center of a circular

park. Angle of elevation of the top of the tower from $A$ and $B$ is ${30^ \circ }$ .

In $\Delta OAC\,\,\tan {30^ \circ } = {h \over a}$

$\Rightarrow {1 \over {\sqrt 3 }} = {h \over a} \Rightarrow h = {a \over {\sqrt 3 }}$