RRB Junior Engineer Stage 1 Model Papers 7

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Question : 22 of 100

Marks: +1, -0

ABCD is a cyclic quadrilateral AB and DC are produced to meet at P. If

∠ADC = 70°

and

∠DAB = 60°

, then the

∠PBC+ ∠PCB

is

130°
150°
155°
180°

Solution:

∠ADC = 70°

∠ABC = 180° – 70° = 110°

⇒ ∠PBC = 70°

∠BCD = 180° – 60° = 120°

⇒ ∠PCB = 60°

∴ ∠PBC + ∠PCB = 70° + 60° = 130°

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