Given a triangle ABC as described below: Side AB = Side AC. Draw a line from C to side AB. call that line CD. Now draw a line from B to side AC. Call that line BE. Let angle EBC = 60 degrees, angle BCD equal 70 degrees, angle ABE equal 20 degrees, and angle DCE equal 10 degrees. Now draw line DE. The question--find what angle EDC is.
(This can be, and is meant to be, solved geometrically, without trig. )
oops i put my tick mark on the wrong side.. giving me the wrong isoceles!! did it again but i still thinkg i got it wrong .. 65? .. if the mid point (let's say O) is where everythign intersects and BOC = 50 thus DOE = 50 .. (my logic might be wrong but i believe triangle DOE is an isoceles) leave 130 degrees divided by 2.. 65? there must be an easier way ahha
EDIT: Vehement answers works out for me.. did that the first time with my wrong isoceles.. but changed my method second time around =/ whoops
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(This can be, and is meant to be, solved geometrically, without trig. )
EDIT: Vehement answers works out for me.. did that the first time with my wrong isoceles.. but changed my method second time around =/ whoops
ok the full steps
first add up bde, ebc, ecb and ecd. you will get 160.
180 - 160 = angel a
180 - angel a(20)/2 = ade = 80
180 - dbe - dbc - bce = angel bdc = 30
subtract bdc(30) and ade(80) from 180 = the answer = 70
thanks much!