Question

In figure, AB is a chord of a circle with centre O and AP is the tangent at A such that $\angle BAP={75}^{\circ }$. Then $\angle ACB$ is equal to:

• A. ${135}^{\circ }$
• B. ${120}^{\circ }$
• C. ${105}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is C ${105}^{\circ }$

Given,

AB is a chord of a circle with centre O and AP is the tangent at A and $\angle BAP={75}^{\circ }$

$\therefore \angle OAP={90}^{\circ }$

$\therefore \angle OAB={90}^{\circ }-{75}^{\circ }$

$={15}^{\circ }$

So, $\angle AOB={180}^0-2\left({15}^0\right)$

$={150}^0$

$\therefore$ Exterior $\angle AOB={360}^0-{150}^0$

$={210}^0$

So, $\angle ACB=\frac{1}{2}\times {210}^0$

$={105}^0$

[Since angle subtended by an arc at the centre is double than the angle subtended by it at the remaining part of the circle]