+0

+1
413
2
+1438

Because I couldn't get any help on the last problem I got a whole bunch more.  Could someone please help with this?

Let AB be a diameter of a circle centered at O. Let E be a point on the circle, and let the tangent at B intersect the tangent at E and AE at C and D, respectively. If angle BAE = 43 degrees, find angle CED, in degrees.

Apr 27, 2018

#1
+101151
+1

Angle BAE  = 43°   ...so   minor  arc BE  will = 86°

And angle CBE  = 1/2 of this arc  = 43°

But.......since  CB and  CE  are tangents  to  the circle from a common point C....they are equal

Thus...in triangle BCE....BC  = EC...and the angles opposite these sides are also equal...

So angle CBE  = angle BEC

And  note that in triangle AEB.....angle AEB  intercepts a diameter....so its measure   = 90°

But  angle AEB  + angle BEC  + angle CED  = 180°.....so....

90   +   43   +   angle CED   = 180

133 + angle  CED  = 180        subtract   133 from each side

angle CED  =  47°

Apr 27, 2018
edited by CPhill  Apr 27, 2018
edited by CPhill  Apr 28, 2018
#2
+1438
+2

Merci beaucoup Monsieur!

AnonymousConfusedGuy  Apr 27, 2018