Question

15. In the given figure, O is centre of the circle. Prove that LOBC + ZBAC = 90°
А
Х
B
C​

Answer 1

Answer:

Step-by-step explanation:

O is the centre of circumscribed circle.

 OB = OC = radii

⇒  ∠OBC = ∠OCB = x

∴ x + x + ∠BOC = 180° (Angle sum property of △OBC)

2x + ∠BOC = 180°

   ∠BOC = 180° - 2x

Also,  ∠BOC = 2∠BAC

  180° - 2x = 2∠BAC

⇒   90 - x = ∠BAC

∴ ∠BAC + ∠OBC = (90° - x) +  x

 ∠BAC + ∠OBC = 90°