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

 Dec 26, 2018
 #1
avatar+101870 
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Because chords CE and BD intersect within the circle, we can use the intersecting chords theorem to find minor arc BC

 

mDQE =  (1/2) ( arc DE + arc BC )

40 = (1/2) ( 30 + arc BC)

80  =  30 + arc BC

50° =  minor arc BC

 

And arc  BAE =

 

360 -  minor arc BC - minor  arc CD -  minor arc DE  =

360 - 50 - 70 - 30 =

360 - ( 150) =

210° 

 

And since PC is a tangent meeting the chord CE....the measure of angle PCE =  (1/2) ( minor arc BC and arc BAE ) = (1/2) ( 50 + 210) = (1/2)(260) = 130°  = mPCQ 

 

And angle CQP = 40°

 

So angle CPQ =  180 - mPCQ - mCQP =  180 -  130 - 40 =  10°

 

And angle CPE is twice this = 20°

 

So....angle CEP =  180 - mPCE - mCPE = 180 - 130 - 20 = 30°

 

And because angle CEP is an inscribed angle, then  minor arc CA is twice this = 60°

 

So

 

arc AE  =  360 -  minor arc CA - minor arc CD - minor arc  DE =

 

360 - 60 - 70 - 30 =

 

360 - [ 60 + 70 + 30 ] =

 

360 - [ 160 ] =

 

200°

 

This does not not look to be possible....I think the problem is that  minor arc DE is given as 30°....however....by the intersecting chords theorem, minor arc BC is computed to be greater than this

But..... minor arc DE looks to be much greater than minor  arc BC (???)

 

 

cool cool cool

 Dec 26, 2018
 #2
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It's just a poorly drawn diagram, (arc CD should be more than double arc DE for example).

I got the same result, via a different route.

If you draw it on paper, using the correct values for the angles, then it all fits together nicely.

 Dec 28, 2018

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