Question

In figure, $PA$ and $PB$ are tangents to the circle with centre $O$ such that $\angle APB=50°$.

Write the measure of $\angle OAB$

Answer 1

Solve for the value of $\angle OAB$

As it is given that PA and PB are two tangents to the circle with center O from point P.

$\Rightarrow PA=PB$

$\Rightarrow \angle PAB=\angle PBA$ { Since $PA=PB$ ,$△PAB$ is an isosceles triangle}

By Angle Sum Property(ASP), sum of angle can be written like

$\begin{array}{rcl}\angle APB+\angle PAB+\angle PBA& =& 180°\\ 50°+\angle PAB+\angle PAB& =& 180°\\ 2\angle PAB& =& 130°\\ \angle PAB& =& 65°\end{array}$

$$\begin{gathered} \Rightarrow \angle OAB=90-\angle PAB=90°-65° \\ \Rightarrow \angle OAB=25° \end{gathered}$$ (the radius from the center of the circle to the point of tangency is perpendicular to the tangent line.)

Hence the value of $\angle OAB$ is $25°$