Question

In the adjoining figure, ABCD is a parallelogram in which $\angle BAD={75}^{\circ }and\angle CBD={60}^{\circ }$. Then, $\angle BDC=$ ?

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

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

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

(d) ${50}^{\circ }$

Answer 1

ANSWER:
(c) 45°
Explanation:
∠B = ${180}^o-\angle A$
⇒ ∠B = ${180}^o-{75}^o={105}^o$
Now, ∠B =​ ∠ABD + ∠CBD
⇒​​ ${105}^o=\angle ABD+{60}^o$
⇒ ∠ABD​ = ${105}^o-{60}^o={45}^o$
⇒ ∠ABD = ​∠ BDC​ = ${45}^o$
(Alternate angles)