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Geometry Problem

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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!

Apr 24, 2018

#1
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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 !!!!]   Apr 25, 2018
#2
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You were right thanks so much!

AnonymousConfusedGuy  Apr 25, 2018