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In triangle ABC, angle ACB is 50 degrees, and angle CBA is 70 degrees. Let D be the foot of the perpendicular from A  to BC, O the center of the circle circumscribed about triangle ABC, and E the other end of the diameter which goes through A. Find the angle DAE, in degrees.

 

 Jun 29, 2018
 #1
avatar+98129 
+2

Extend AD through to intersect the circle at F

 

Now.....since angle ACB  =   50°...then minor arc AB has twice its measure  =100°

 

And since angle  CBA  =70°....then  angle BAD  = BDA  - CBA  = 90 - 70  = 20° = angle BAF

 

And angle BAF  =  intercepts minor arc BF...so  the measure of minor arc BF  is twice this = 40°

 

But arc ABE  =180°   =  minor arc AB  + minor arc BF  + minor arc FE

 

So

 

180  =  100 + 40  +  minor arc FE

 

180  =140 + minor arc FE

 

40  = minor arc FE

 

But angle FAE  intercepts minor arc FE, so it has  1/2 of its measure =  20°

 

But angle  FAE  = angle DAE...so  DAE  = 20°

 

 

 

cool cool cool

 Jun 29, 2018
 #2
avatar+7350 
+2

Also....

 

 

 

There are  180°  in triangle ACD, so..

m∠DAC  =  180° - 90° - 50°

m∠DAC =  40°

 

An inscribed angle is half the measure of its intercepted arc, so....

(1/2)m∠AOC  =  m∠ABC

(1/2)m∠AOC  =  70°

m∠AOC  =  140°

 

OA  and  OC  are radii of circle  O , so they are the same length, and triangle AOC  is isosceles. So...

m∠OAC  =  m∠OCA

m∠OAC + m∠OCA + 140°  =  180°

m∠OAC + m∠OAC + 140°  =  180°

2m∠OAC + 140°  =  180°

2m∠OAC  =  40°

m∠OAC  =  20°

 

m∠DAE  =  m∠DAC - m∠OAC

m∠DAE  =  40° - 20°

m∠DAE  =  20°

 Jun 29, 2018
 #3
avatar+98129 
+1

I like yours better, hectictar....no extraneous lines need be drawn....!!!!

 

 

cool cool cool

CPhill  Jun 29, 2018

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