In a circle with centre O, PA and PB are tangents to the circle at point A and point B, respectively. C is a point on the major arc AB. If ∠ACB= 50°, then find the measure of ∠APB.

80°

According to the concept,

⇒ \(\angle\)AOB = 2 x \(\angle\)ACB

⇒ \(\angle\)AOB = 2 x \({50}^\circ\)

⇒ \(\angle\)AOB = \({100}^\circ\)

Consider the quadrilateral AOBP,

⇒ \(\angle\)A + \(\angle\)O + \(\angle\)B + \(\angle\)P = \({360}^\circ\)

⇒ \({90}^\circ\) + \({100}^\circ\) + \({90}^\circ\) + \(\angle\)P = \({360}^\circ\)

⇒ \(\angle\)P = \({360}^\circ\) - \({90}^\circ\) - \({90}^\circ\) - \({100}^\circ\)

⇒ \(\angle\)P = \({80}^\circ\)

⇒ \(\angle\)APB = \({80}^\circ\)

Therefore, \(\angle\)APB is \({80}^\circ\).