Question 36
ABCD is a cyclic quadrilateral. AB and DC are produced to meet at P. If angleADC = $$70°$$ and angleDAB = $$60°$$, then the $$\angle{PBC} + \angle{PCB}$$ is
Solution
$$\because$$ ABCD is cyclic quadrilateral.
$$\therefore \angle{DAB}+\angle{DCA}=180$$
$$\angle{DCA}=120$$
Similarly,$$\angle{ABC}=110$$
$$\angle{PBC}=180-\angle{ABC}$$
$$\angle{PBC}=70$$
Similarly,$$\angle{PCB}=60$$
$$\therefore \angle{PBC}+\angle{PCB}=70+60=130$$
Hence, Option A is correct.
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