Question

In the given figure, ABCD is a parallelogram in which $\angle BDC={45}^{\circ }and\angle BAD={75}^{\circ }$. Then, $\angle CBD=$ ?

(a) ${45}^{\circ }$

(b) ${55}^{\circ }$

(c) ${60}^{\circ }$

(d) ${75}^{\circ }$

Answer 1

ANSWER:
(c)​ 60°
Explanation:
∠BAD = ∠BCD = ${75}^{\circ }$
[Opposite angles are equal]
In ∆ BCD, ∠ C = ${75}^{\circ }$
∴ ​∠ CBD = ${180}^{\circ }$ − (${75}^{\circ }$ + ${45}^{\circ }$ ) = ${60}^{\circ }$