Question

In the adjoining figure, O is the centre of the circle. Determine $\angle$ACB, if PA and PB are tangents and $\angle$APB$={60}^o$.

Answer 1

In Quadrate AOBP

Angle sum = 360

$\Rightarrow \angle A+\angle B+\angle P+\angle 0=360$

$90+90+60+10=360$

$\Rightarrow \angle 0={120}^{\circ }$

In circle, Angle subtend by minor segment center

is twice as Angle subtend are major segment

$\Rightarrow \angle ACB=\frac{1}{2}\angle AOB$

$\Rightarrow \angle ACB={60}^{\circ }$