Question

Vertex C, of $\angle PCQ=x$ lies on the circle. The arms of $\angle PCQ$ intersect the circle at A and B. If m(arc AB)= 120${}^{\circ }$, then $x$=

• A. 150${}^{\circ }$
• B. 120${}^{\circ }$
• C. 60${}^{\circ }$
• D. 30${}^{\circ }$

Answer 1

The correct option is C 60${}^{\circ }$


By inscribed angle theorem, we know that the measure of an angle subtended by an arc at a point on the circle is half of the measure of the angle subtended by the arc at the centre.
Given, m(arc AB) = 120${}^{\circ }$
Then, $\angle PCQ=\angle ACB=x=\frac{120}{2}={60}^{\circ }$