Question

From an external point $P$, tangents $PA$ and $PB$ are drawn to a circle with centre $O$. If $\angle PAB={50}^{\circ },$ then find $\angle AOB$.

Answer 1

Given, $PA=PB$, and $\angle PAB={50}^o$

To prove: Measure of $\angle AOB$

Proof: $PA=PB$ ($\because$ given)

$\angle PBA=\angle PAB$ ($\because$ angle opposite sides are equal)

$\therefore \angle PBA={50}^o$.

In $\Delta PAB$

$\angle PBA+\angle PAB+\angle APB={180}^o$ ($\because$ A.S.P)

$\angle APB={180}^o-{100}^o$

$\angle APB={80}^o$

$\angle AOB+\angle APB={180}^o$ ($\therefore$ supplementary angles)

$\angle AOB={180}^o-{80}^o$

$\angle AOB={100}^o$

$\therefore$ Hence proved.