A tower stands at the centre of a circular park.

Question

A tower stands at the centre of a circular park. $A$ and $B$ are two points on the boundary of the park such that $AB(=a)$ subtends an angle of $${60^ \circ }$$ at the foot of the tower, and the angle of elevation of the top of the tower from $A$ or $B$ is $${30^ \circ }$$. The height of the tower is :

Accepted Answer

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 }}$

Solutions · 1