+1452 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!
+129928 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 !!!!]
