Question

Two tangents $PA$ and $PB$ are drawn from an external point $P$ to a circle inclined to each other at an angle of ${70}^o$, then find $\angle AOP$.

Answer 1

$\angle APB={70}^o$

$\therefore \angle APO=\frac{1}{2}\angle APB$

$=\frac{1}{2}\times 70$

$={35}^o$

$\angle PAO={90}^o$

$\therefore$ In $△OAP$

$\angle OAP+\angle APO+\angle POA={180}^o$

$\angle AOP={180}^o-35-90$

$\angle AOP={55}^o$