Question

In the given figure , AC and BC are tangents to the circle with centre O.

If $\angle ACB$ is twice $\angle AOB$ , then find the $\angle ACB$.

Answer 1

$\angle OAB$ = $\angle OBC$ = ${90}^{\circ }$ ...........(Tangent is perpendicular to the radius at the point of contact.)

$\angle AOB$ + $\angle ACB$ = ${180}^{\circ }$ ...........(Sum of angles of a quadrilateral)

But, $\angle ACB$ = 2$\angle AOB$ .........(Given)

$\angle AOB$ + 2$\angle AOB$ = ${180}^{\circ }$

3$\angle AOB$ = ${180}^{\circ }$

$\angle AOB$ = ${60}^{\circ }$

$\therefore \angle AOB$ = 2$\times$${180}^{\circ }$ = ${120}^{\circ }$