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In the diagram shown, PC is tangent to the circle and PD is the angle bisector of CPE. If arc CD is 70 degrees, arc DE is 30 degrees, and DQE = 40 degrees, then determine arc AE, in degrees.

Thanks!

AnonymousConfusedGuy Apr 24, 2018

#1**+1 **

Because CE is a chord meeting a tangent, the measure of the angle supplemental to angle PCQ =

(1/2) ( m arc CDE) = (1/2) (m arc CD + m arc DE ) = (1/2) ( 70 + 30) = 50°

So....angle PCQ = 180 - 50 = 130°

And angle CQP is vertical to angle DQE = 40°

So....in triangle PCQ, angle PCQ = 130 and angle CQP = 40

So angle CPQ = 180 - 130 - 40 = 10°

And angle QPE equals this = 10°

And angle PQE is supplemental to angle DQE = 140°

So...in triangle PQE, angle QEP = 180 - 140 - 10 = 30°

So m arc CA is twice this = 60°

So m arc AE = 360 - m arc DE - m arc CD - m arc CA =

360 - 30 - 70 - 60 =

360 - 160 =

200°

This doesn't "look " to be correct......[ I don't think I made any mistakes, but maybe I did !!!!]

CPhill Apr 25, 2018