Question

An isosceles triangle ABC is inscribed in a circle with centre O. If $\angle BOC$ is right angle, then $\angle ABC$ is equal to ___. (Given : $AB=AC$)

• A. 67.5${}^{\circ }$
• B. 45${}^{\circ }$
• C. 60${}^{\circ }$
• D. 30${}^{\circ }$

Answer 1

The correct option is A 67.5${}^{\circ }$

Given that $\Delta ABC$ is an isosceles.
$AB=AC$
$⟹\angle ABC=\angle ACB$ ...(1)
$⟹\angle BOC={90}^{\circ }$
Since the angle subtended by a chord at the centre of a circle is twice the angle subtended by the same chord at any other point on the remaining part of the circle, we have
$\angle BOC=2\angle BAC$
$\angle BAC=\frac{1}{2}$×${90}^{\circ }$
$\angle BAC={45}^{\circ }$ ...(2)
Consider $\Delta ABC$.
$\angle BAC+\angle ABC+\angle ACB={180}^{\circ }$ ...(3)
From equations (1), (2) and (3),
$2\angle ABC={135}^{\circ }$
$\angle ABC={67.5}^{\circ }$