Question

In ΔABC, start overline, A, C, end overline
AC
is extended through point C to point D, m, angle, B, C, D, equals, left parenthesis, 9, x, plus, 2, right parenthesis, degreesm∠BCD=(9x+2)
∘
, m, angle, C, A, B, equals, left parenthesis, 3, x, plus, 20, right parenthesis, degreesm∠CAB=(3x+20)
∘
, and m, angle, A, B, C, equals, left parenthesis, x, plus, 17, right parenthesis, degreesm∠ABC=(x+17)
∘
. Find m, angle, B, C, D, .m∠BCD.

Answer 1

Because C is on the extension of AC, ∠BCD is an exterior angle of ΔABC, so
m∠BCD = m∠CAB + m∠ABC.

Thus 9x + 2 = (3x + 20) + (x + 17) = 4x + 37.
Solve: 9x + 2 = 4x + 37 → 5x = 35 → x = 7.

Then m∠BCD = 9x + 2 = 9(7) + 2 = 65°.
Answer: 65°.