Question

${}^{\prime }{O}^{\prime }$ is a centre of a circle, $AC$ is a tangent to a circle at point $A$. If $△OAC$ is an isosceles triangle, then find the measure of $\angle OCA$

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. ${90}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is D ${45}^{\circ }$

Given that $O$ is center of circle and $AC$ is tangent to circle at $A$

Given that triangle $AOC$ is isosceles.

We know that angle $OAC={90}^0$

So sum of other two angles in triangle $OAC$ must be ${90}^0$

Since the triangle is isosceles, the two angles must be equal.

So we get angle $OCA=\frac{{90}^0}{2}={45}^0$

Hence, the correct option is $D$