Question

In given figure $C$ is centre of circle. $AO$ and $BO$ are tangents to circle. $CM\perp AB$.

If $AC=AB=6cm$, then find $\angle AOB$.

• A. ${90}^o$
• B. ${120}^o$
• C. ${145}^o$
• D. ${150}^o$

Answer 1

Answer: (B)

${120}^o$

C is the center of the circle.

$\therefore CA=CB=$ radii of the circle

$\therefore △ABC$ is isosceles.

But, $CA=AB$ ....given

$\therefore △ABC$ is an equilateral triangle.

$\therefore \angle ACB={60}^o$

In $\square ACBO$,

$\angle A+\angle AOB+\angle B+\angle ACB={360}^o$

$\therefore \angle AOB={360}^o-{90}^o-{90}^o-{60}^o={120}^o$.