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 - $$\angle DPC =$$

Solution

In $$\triangle ABP$$,

$$\angle A + \angle ABC + \angle APB = 180\degree$$

$$\angle APB = 180 - 60 - 72 = 48\degree $$

$$\angle ADC = 180 - ABC = 180 - 72 = 108\degree$$
$$\angle PDC = 180 - \angle ADC = 180 - 108 = 72\degree$$

$$\angle PDC - \angle DPC = 72 - 48 = 24\degree$$

Share on Whatsapp Share on Whatsapp

Create a FREE account and get: