Question

A triangle is formed with all its vertices lying on a circle. One of the sides of the triangle is equal to the radius of the circle. Find the angle of the triangle opposite to the side having a length equal to the circle's radius.

degrees

Answer 1

Draw a triangle $△ABC,$ with all its vertices lying on a circle having its centre at O.

The schematic is as follows:

Here, $OB=OC=BC$ = radius of the circle
Hence, $△BOC$ is an equilateral triangle.
$\therefore \angle BOC={60}^o$

We know, the angle subtended by an arc at the centre of a cirlce is double the angle subtended by the arc at any point on the circle.

$\therefore \angle BOC=2\times \angle BAC$
$\Rightarrow \angle BAC=\frac{{60}^o}{2}={30}^o$
Also, the angle of triangle $△ABC$ opposite to the side having a length equal to the circle's radius is $\angle BAC.$

Hence, the required angle is ${30}^o.$