Question

In the given figure, ∠BAC = 30 °, ∠ABC is = 50° and ∠CDE = 40°. Find ∠AED.

Answer 1

in triangle abc.
angle BAC is equal to 30 degree ,,,,angle abc is equal to 50 degree.
then angle ACB is equal to.
angle BAC + angle ACB + angle abc= 180.
( by angle sum property)
30°+ angle ACB+50°=180°
80°+ angle ACB=180°
angle ACB=100°.
since BCD is a straight line sum 180° .
in triangle in ECD.
angle EDC=40° .
angle ECD.is equal to 80° .
then angle CED is equal to 60 degree.
since a AEC is also a straight line.
then angle AED + angle CED =180° .
then angle AED =120° .
hope it helps.
^_^ .

Answer 2

Answer is

Angle AED = 120 degree