Geometry Practice Test 7

In ΔACB,
AC and BC are two tangents to two circles.

∠CAB = ∠CBA = x°
(angles with tangent at outer points are drawn equal.)
∠ACB + ∠CAB + ∠CBA = 180°
45° + x + x = 180°
2x = 135
x = 67.5°
∠CAB = ∠AQP = 67.5° (alternative interior segment property)
∠CBA = ∠BQP = 67.5°
∠AQP + ∠BQP = ∠AQB
∠AQB = 135°