Question

In the given figure, PA and PB are tangents to the circle with centre O such that $\angle \mathrm{APB}=50°$. Write the measure of $\angle \mathrm{OAB}.$ [CBSE 2015]

Answer 1

It is given that tangents PA and PB are drawn from an external point P to a circle with centre O.

∴ PA = PB (Lengths of tangents drawn from an external point to a circle are equal)

In ∆PAB,

PA = PB

∴ $\angle \mathrm{PBA}=\angle \mathrm{PAB}$ (In a triangle, equal sides have equal angles opposite to them)

Now,

$\angle \mathrm{PAB}+\angle \mathrm{PBA}+\angle \mathrm{APB}=180°$ (Angle sum property)
$$\begin{gathered} \Rightarrow 2\angle \mathrm{PAB}+50°=180°\left(\angle \mathrm{APB}=50°\right) \\ \Rightarrow 2\angle \mathrm{PAB}=180°-50°=130° \\ \Rightarrow \angle \mathrm{PAB}=65° \\ \therefore \angle \mathrm{PBA}=\angle \mathrm{PAB}=65°.....\left(1\right) \end{gathered}$$

Now, PA is the tangent and OA is the radius through the point of contact A.

∴ $\angle \mathrm{OAP}=90°$ (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

Now,

$\angle \mathrm{OAB}=\angle \mathrm{OAP}-\angle \mathrm{PAB}=90°-65°=25°\left[\mathrm{Using}\left(1\right)\right]$

Hence, the measure of $\angle \mathrm{OAB}$ is 25º.