Question 66
ABCD is a cyclic quadrilateral which sides AD and BC are produced to meet at P, and sides DC and AB meet at Q when produced. If $$\angle A = 60^\circ$$ and $$\angle ABC = 72^\circ$$, then \angle PDC -Â $$\angle DPC =$$
Solution
In $$\triangle ABP$$,
$$\angle A + \angle ABC + \angle APB = 180\degree$$
$$\angle APB = 180 - 60 - 72 = 48\degree $$
$$\angle ADC = 180 - ABC = 180 - 72 = 108\degree$$
$$\angle PDC = 180 -Â \angle ADC = 180 - 108 = 72\degree$$
$$\angle PDC - \angle DPC = 72 - 48 = 24\degree$$
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