Q9 of 100 Page 17

In Fig. 16.200, P and Q are centres of two circles intersecting at B and C. ACD is a straight line. Then, ∠BQD=

We know that,

∠ACB =


∠ACB =


∠ACB = 75o


Since,


ACD is a straight line, so


∠ACB + ∠BCD = 180o


75o + ∠BCD = 180o


∠BCD = 180o – 75o


= 105o


Now,


∠BCD = Reflex ∠BQD


105o = (360o - ∠BQD)


210o = 360o - ∠BQD


∠BQD = 360o – 210o


Therefore,


∠BQD = 150o


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