Question

ABCD is a parallelogram and DM ⊥ BM as shown in the figure. Find $\angle CDM$ (in degrees).

• A. 60

Answer 1

The correct option is A 60
ABCD is a parallelogram with $\angle ADC={30}^{\circ }.$

$DM\perp BM$

Here, AD || BC
(Since opp. sides of a parallelogram are equal)

Since the perpendicular drawn from point D meets BC at point M, AD || BC || BM.

$\Rightarrow \angle ADC=\angle DCM={30}^{\circ }$
(Since alternate interior angles are equal)

In $△DCM$,

$\angle CDM+\angle DMC+\angle DCM={180}^{\circ }$
(using angle sum property)

$\Rightarrow \angle CDM+{90}^{\circ }+{30}^{\circ }={180}^{\circ }$

$\Rightarrow \angle CDM={60}^{\circ }$