AB is the diameter of a circle with centre O . If P be a point on the circle such that $\angle A O P=110^{\circ}$, then the measure of $\angle O B P$ is:
Answer & explanation
Correct answer: option 4
⇒ OA = OP = OB = radius
In \(\Delta \)OPB
⇒ OP = OB
so, \(\Delta \)OPB is an isosceles triangle
⇒ $\angle O P B$ = $\angle O B P$ = x
As, AB is an straight line
⇒ \(\angle\)AOP + \(\angle\)POB = \({180}^\circ\)
⇒ \(\angle\)POB = \({108}^\circ\) - \({110}^\circ\) = \({70}^\circ\)
In \(\Delta \)OPB
⇒ \(\angle\)O + ⇒ \(\angle\)P + ⇒ \(\angle\)B = \({180}^\circ\)
⇒ \({70}^\circ\) + x + x = \({180}^\circ\)
⇒ 2x = \({180}^\circ\) - \({70}^\circ\) = \({110}^\circ\)
⇒ x = \({55}^\circ\)
Therefore, $\angle O P B$ is \({55}^\circ\)