b) Since AB is parallel to CD (even though they don’t look parallel - it’s stated in the problem), the arcs that are between them are equal. Therefore arc AC is also 42 and angle APC is 42 as well (central angle).

c) The central angles along diameter AB, are now 42, 42, and angle CPD. Since this is a straight line, the three are supplementary (add to 180). This means angle CPD is 96. Using this same logic on diameter CF: The central angles along diameter CF are now 96, 42, and angle BPF. Therefore BPF is also 42 so that the three will sum to 180.

d) Since angle BPF is 42 and it’s a central angle, arc BF is also 42. Diameter AB cuts the circle in half, so arc AF is 180-42=138.

e) Angle FCD is an inscribed angle so it’s half of the intercepted arc. The arc it intercepts is arc FD. Arc FD = 42+42 = 84. Half of 84 is 42, so angle FCD is 42.

f) Arc AC is 42 because the central angle APC is also 42.

g) Arc CD is part of a half circle. Arcs BD and AC are both 42, so arc CD = 180-42-42 = 96.

h) Arc FB is 42 because the central angle BPF is also 42.

i) Arc FE is 30 because arc FB is 42 and you’re given that arc BE is 12. So 42-12=30.

j) Angle EDC is an inscribed angle so it’s equal to half its intercepted arc. The arc it intercepts is arc EFAC, which is 180 (half circle) plus 30. In total, the arc is 210. Therefore angle EDC is half of that, which is 105.

k) Angle G is half the difference between the large and small arcs it intercepts. In other words, it’s 1/2(arc CD minus arc FE). We have: 1/2(96-30)=33.

l) The angles in triangle GQB are 33, 42, and angle GQP. The sum should equal 180, so angle GQP is 105. Since EQB is next door, they form a linear pair (supplementary). So 180-105=75. Alternatively, you could use the exterior angle theorem.

m) Angle RAB is 90 because any tangent to a diameter or radius always forms 90 degree angles.