Question

In ΔJKL, start overline, J, L, end overline
JL
is extended through point L to point M, m, angle, L, J, K, equals, left parenthesis, 3, x, plus, 13, right parenthesis, degreesm∠LJK=(3x+13)
∘
, m, angle, K, L, M, equals, left parenthesis, 8, x, minus, 13, right parenthesis, degreesm∠KLM=(8x−13)
∘
, and m, angle, J, K, L, equals, left parenthesis, 2, x, minus, 8, right parenthesis, degreesm∠JKL=(2x−8)
∘
. Find m, angle, K, L, M, .m∠KLM.

Answer 1

Because M is on the extension of JL, ∠KLM is an exterior angle of ΔJKL, so
m∠KLM = m∠LJK + m∠JKL.

Substitute expressions:
8x − 13 = (3x + 13) + (2x − 8) = 5x + 5.

Solve: 8x − 13 = 5x + 5 → 3x = 18 → x = 6.

Then m∠KLM = 8x − 13 = 8(6) − 13 = 48 − 13 = 35°.

Answer: 35°.