Vertices A, B, C and D of a quadrilateral ABCD lie on a circle. ∠A is three times ∠C and ∠D is two times ∠B. What is the difference between the measures of ∠D and ∠C ?
Answer & explanation
Correct answer: option 3
= \(\angle\)A + \(\angle\)C = \({180}^\circ\)
= 3\(\theta \) + \(\theta \) = \({180}^\circ\)
= \(\theta \) = \({45}^\circ\)
= \(\angle\)C = \({45}^\circ\)
= \(\angle\)B + \(\angle\)D = \({180}^\circ\)
= \(\alpha \) + 2\(\alpha \) = \({180}^\circ\)
= \(\alpha \) = \({60}^\circ\)
= \(\angle\)B = \({60}^\circ\)
The difference between the measure of \(\angle\)D and \(\angle\)C = (120 - 45) = \({75}^\circ\)