Practicing Success
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: |
50° 65° 60° 55° |
55° |
⇒ 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 \({55}^\circ\) |