If y=15, then what is \angle ACB equal to?

Consider the following for the next two (02) items that follow:<br>In the following figure, a triangle ABC is inscribed in a circle with centre at O. Let \angle POA=x^{\circ} and \angle OQB=y^{\circ}. Further, OB = BQ.

  1. A. 30^{\circ}
  2. B. 40^{\circ}
  3. C. 45^{\circ}
  4. D. 60^{\circ}

Correct Answer: D. 60^{\circ}

Explanation

For y=15, the central angle \angle AOB = 180^{\circ} - (\angle OAB + \angle OBA) = 180^{\circ} - 4y = 120^{\circ}. The inscribed angle \angle ACB subtending the same arc is half of the central angle, so \angle ACB = \frac{120^{\circ}}{2} = 60^{\circ}.

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