Question

A circle (with centre at O) is touching two intersecting lines AX and BY. The two points of contact A and B subtend an angle of ${65}^{\circ }$ at any point C on the circumference of the circle. If P is the point of intersection of the two lines, then the measure of $\angle$ APB is

• A. ${50}^{\circ }$
• B. ${65}^{\circ }$
• C. ${90}^{\circ }$
• D. ${40}^{\circ }$

Answer 1

The correct option is A ${50}^{\circ }$

$\angle AOB=2\angle ACB$ (Angle subtended by an arc at the centre is double the angle subtended by it at the circumference)

$⟹\angle AOB=2\times 65={130}^{\circ }$

In quadrilateral $AOBP$
$\angle OAP+\angle OBP+\angle AOB+\angle APB={360}^{\circ }$
$⟹{90}^{\circ }+{90}^{\circ }+\angle APB+{130}^{\circ }={360}^{\circ }$
$⟹\angle APB={360}^{\circ }-{310}^{\circ }={50}^{\circ }$