Given that ∠XQR = 180° and ∠LQM = 180°, which equation could be used to solve problems involving the relationships between ∠XQM and ∠RQM?

Responses

A 180 + (3a + 39) = (136 − 2a)180 + (3a + 39) = (136 − 2a)

B (3a + 39) − 180 = (136 − 2a)(3a + 39) − 180 = (136 − 2a)

C (136 − 2a) − (3a + 39) = 180(136 − 2a) − (3a + 39) = 180

D (136 − 2a) + (3a + 39) = 180(136 − 2a) + (3a + 39) = 180

E 360 − (3a + 39) = (136 − 2a)

The correct answer is C. (136 − 2a) − (3a + 39) = 180. This equation represents the relationship between angles ∠XQM and ∠RQM, where (136 − 2a) represents ∠XQM and (3a + 39) represents ∠RQM.