Question

Given: Triangles ABC and DBC are isosceles, m∠BDC = 30°, and m∠ABD = 155°. Find m∠ABC, m∠BAC, and m∠DBC.

Answer 1

∠BDC=30, so ∠CBD + ∠BCD=180-30=150
ΔDBC is isosceles, so ∠CBD=∠BCD=half of 150=75
∠ABC=∠ABD-∠CBD=155-75=80
ΔABC is isosceles, so ∠ABC=∠ACB=80
∠BAC=180-80-80=20