∆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°
#4
(5/4)*5= (25/4)=6 1/4 miles per hr
#5
(40/$62) = (x/$128.65)
x = 83
The answer is 67°.
Since the parallel sides of a trapezoid are called bases, the bases in our given trapezoid are AB and DC. We know that the corresponding pairs of base angles, such as ∠A and ∠D, or ∠B and ∠C, are supplementary, therefore, their angles add up to 180 degrees:
∠A + ∠D = 180°
∠B + ∠C = 180°
Given that m∠A = 113°, we can calculate for the measure of ∠D:
113° + ∠D = 180°
∠D = 180° - 113° = 67°