In Fig. 16.199, AB is a diameter of the circle such that ∠A=35° and ∠Q=25°, find ∠PBR.
In triangle ABQ,
∠ABQ + ∠AQB + ∠BAQ = 180o
∠ABQ + 25o + 35o = 180o
∠ABQ = 120o
∠APB = 90o (Angle in the semi-circle)
In triangle APB,
∠APB + ∠PBA + ∠PAB = 180o
90o + ∠PBA + 35o = 180o
∠PBA = 55o
Now,
∠PBR = ∠PBA + ∠PBR
∠PBR = 55o + (180o – 120o)
∠PBR = 115o
Thus,
∠PBR = 115o