First problem.
m<CKL = m<CJI = y
m<CLM = m<CFG
8x = 42 + y
7x = 25 + y
x = 17
m<CAB = m<CJK = 180 - y
7(17) = 25 + y
y = 94
m<CAB = m<CJK = 180 - 94 = 86
m<CAB = 86
m<CBA = m<CAB
m<CBA = 86
One interior angle is 30.
One exterior angle is 40.
180 - 40 = 140
The interior angle at the vertex that has the 40-deg exterior angle is 140 deg.
180 - 140 - 30 = 10
Angles: 10, 140 deg
Problem 21a.
m<DBA = m<CBE by Vertical angles theorem
Problem 21b.
m<ACD = m<BCD by Corresponding parts of congruent triangles are congruent (CPCTC)
