UPSC CDS 1 2023 Math Paper

Concept:Use the exterior angle theorem and the central angle theorem to relate angles in the circle.Explanation:In triangle $OBQ$, we have $OB = BQ$ and $\angle OQB = y = 15^{\circ}$.Since $OB = BQ$, the base angles are equal: $\angle BOQ = \angle BQO = 15^{\circ}$.The exterior angle at $B$ in triangle $OBQ$ is $\angle OBA = \angle BQO + \angle BOQ = 15^{\circ} + 15^{\circ} = 30^{\circ}$.$OA = OB$ (radii of the same circle), so triangle $OAB$ is isosceles: $\angle OAB = \angle OBA = 30^{\circ}$.In triangle $AOB$, the sum of angles is $180^{\circ}$, so $\angle AOB = 180^{\circ} - 30^{\circ} - 30^{\circ} = 120^{\circ}$.By the central angle theorem, the angle at the center ($\angle AOB$) is twice the angle at the circumference ($\angle ACB$) standing on the same arc $AB$: $\angle AOB = 2\angle ACB$.Thus, $\angle ACB = \frac{120^{\circ}}{2} = 60^{\circ}$.Answer:$\angle ACB = 60^{\circ}$ (Option D).