Question

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

Answer 1

It is given that PA and PB are tangents to the given circle.
$\therefore \angle \mathrm{PAO}=90°$ (Radius is perpendicular to the tangent at the point of contact.)
Now,
$\angle \mathrm{PAB}=50°$ (Given)
$\therefore \angle \mathrm{OAB}=\angle \mathrm{PAO}-\angle \mathrm{PAB}=90°-50°=40°$
In ∆OAB,
OB = OA (Radii of the circle)
$\therefore \angle \mathrm{OAB}=\angle \mathrm{OBA}=40°$ (Angles opposite to equal sides are equal.)
Now,
$\angle \mathrm{AOB}+\angle \mathrm{OAB}+\angle \mathrm{OBA}=180°$ (Angle sum property)
$\Rightarrow \angle \mathrm{AOB}=180°-40°-40°=100°$