Question

The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O such that $\angle DAC={30}^{\circ }and\angle AOB={70}^{\circ }$. Then, $\angle DBC=$ ?

(a) ${40}^{\circ }$

(b) ${35}^{\circ }$

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

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

Answer 1

ANSWER:
(a) 40°
Explanation:
∠ OAD = ​∠ OCB = ${30}^{\circ }$
(Alternate interior angles)
∠ AOB + ∠ BOC = ${180}^{\circ }$
(Linear pair of angles)
∴ ∠ BOC =${180}^{\circ }-{70}^{\circ }={110}^{\circ }$
(∠ AOB = ${70}^{\circ }$)
In ∆ BOC, we have:
∠ OBC = ${180}^{\circ }-({110}^{\circ }+{30}^{\circ })={40}^{\circ }$
∴ ​∠ DBC = ${40}^{\circ }$