Question

In the given figure ABC is a triangle in which angle BAC is equal to 30 degree show that BC = the radius of the circumcircle of triangle ABC whose Centre is O. ​

Answer 1

See the figure attached..

Explanation:

First, Join OB and OC.

∠BOC = 2∠BAC

(∵ angle subtended by an arc at the centre = double the angle subtended by it at any point on the remaining part of the circle)

=> ∠BOC = 2 x 30°...(∵ ∠BAC = 30°, Given)

=> ∠BOC = 60°

In, ΔOBC, OB = OC...(radii of same circle)

∴ ∠OCB = ∠OBC...(∠s opp. equal sides of a triangle are equal)

As the sum of angles of a traingle is 180°,

∠OBC + ∠OCB + ∠BOC = 180°

∠OBC + ∠OBC + 60° = 180°...(∵∠BOC = 60°,proved above)

2∠OBC = 120° => ∠OBC = 60°

Thus, we have ΔABC such that,

∠BOC = ∠OBC = ∠OCB = 60°

ΔOBC is an equilateral triangle

BC = OB i.e. BC is equal to the radius of circle.

Hence proved...

Hope it helps you!!