Q. 85.0( 1 Vote )

In Fig. 16.199, <

Answer :

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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