Q8 of 100 Page 17

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


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