Q19 of 109 Page 318

In the given figure, ABCD is a parallelogram in which ∠BDC = 45° and ∠BAD = 75°. Then, ∠CBD = ?

As per the question,

∠ BAD = ∠ BCD = 75° (opposite angles of parallelogram)

Now, in ΔBCD,

∠BCD + ∠CBD + ∠BCD = 180°

45 + ∠CBD + 75 = 180°

∠CBD = 60°

∴ Option C is correct

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