Question

In fig 9.7, if PA and PB are tangents to the circle with centre O such that $\angle APB={50}^0$, then $\angle OAB$ is equal to

• A. ${25}^0$
• B. ${30}^0$
• C. ${40}^0$
• D. ${50}^0$

Answer 1

The correct option is A ${25}^0$

as we know that tangents drawn from a point to a circle are equal therefore

$PA=PB$ and so $\angle PAB=\angle PBA$

$\angle PAB+\angle ABP+\angle BPA=180$

$\angle PAB=\frac{180-50}{2}=65$

therefore

$\angle OAB=90-65=25$[since PA is perpendicular to AO]