heeeeeeeelp math
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?

Given ΔABC,
A=20°
B=80° =>
C=80°
=> isosceles triangle with base BC=x.
Let E=midpoint of BC, then
ΔAEC is a right triangle rightangled at E
EC=x/2
By definition of cosine,
AC=(x/2)/cos(80°)
=x/(2cos(80°))
Consider ΔADC,
AD=x (given)
AC=x/(2cos(80°)) (from above)
∠DAC=20° (given),
we find DC by cosine rule
DC=sqrt(AD²+AC²2*AD*AC*cos(20°) )
=sqrt(x²+x/(2cos(80°))x²cos(80°))
=1.879x (approx.)
∠ ADC can be found by the sine rule:
sin(ADC)=(x/(2cos(80°))*sin(20°)/DC
=sin(20°)/(12cos(80°))
∠ADC=asin(sin(20°)/(12cos(80°)))
=31.6° approx.
Please check my arithmetic.