posted by lin .
In triangle ABC, ∠A=20∘ and ∠B=80∘. Let
D be a point on line segment AB such that
AD=BC. What is the measure (in degrees) of ∠ADC?
=> isosceles triangle with base BC=x.
Let E=mid-point of BC, then
ΔAEC is a right triangle right-angled at E
By definition of cosine,
AC=x/(2cos(80°)) (from above)
we find DC by cosine rule
∠ ADC can be found by the sine rule:
Please check my arithmetic.