Question

Help with the problem pls

Answer 1

∆AQB is isosceles, so ∠QAB = ∠QBA.

The sum of vertex angles of ∆ABC is 180° so you have

... ∠QAB + ∠QCB + (∠QBC + ∠QBA) = 180°

Substituting what we know ...

... ∠QAB + 52° + (28° + ∠QAB) + 180°

... 2∠QAB = 100°

... ∠QAB = 50°

Because PQ is parallel to BC, ∠APQ = ∠ABC = 50° + 28°

... ∠APQ = 78°