In triangle ABC, ∠A=20∘ and ∠B=80∘. Let

Question

D be a point on line segment AB such that
AD=BC. What is the measure (in degrees) of ∠ADC?

Answer 1

Given ΔABC,
A=20°
B=80° =>
C=80°
=> isosceles triangle with base BC=x.

Let E=mid-point of BC, then
ΔAEC is a right triangle right-angled 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+AC²-2*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°)/(1-2cos(80°))

∠ADC=asin(sin(20°)/(1-2cos(80°)))
=31.6° approx.

Please check my arithmetic.