Question

The centre of circumcircle of abc is o. prove that obc+bac=90*

Answer 1

Let

angle OBC = x°

angle BAC = y°

Solution:-

In ∆OBC,

OB = OC (radius of circle)

=> angle OBC = angle OCB = x°

angle OCB = x°

Taking chord BC,

angle BOC = 2 × angle BAC (angle subtended at centre is twice the angle at circumference)

=> angle BOC = 2y°

In ∆OBC,

angle OBC + angle OCB + angle BOC = 180° (angle sum property)

=> x° + x° + 2y° = 180°

=> 2x° + 2y° = 180°

Taking ‘2’ as common on both sides,

=> 2(x° + y°) = 2(90°)

Dividing both sides by 2,

=> x° + y° = 90°

W.K.T x° = angle OBC , y° = angle BAC

Therefore, angle OBC + angle BAC = 90°