Question

Given a parallelogram ABCD, if $\angle A={60}^{\circ }$ then $\angle D$ is equal to :

• A. ${110}^{\circ }$
• B. ${140}^{\circ }$
• C. ${120}^{\circ }$
• D. ${130}^{\circ }$

Answer 1

The correct option is C ${120}^{\circ }$

$\angle A={60}^o,$ therefore the opposite angle $C={60}^o$

$\angle B=\angle D$ [Rule of parallelogram] as they are opposite angles.

Let $\angle B$ and $\angle D$ be $x$

$\Rightarrow x+x+\angle A+\angle B={360}^o$ [ sum of all sides of a polygon ]

$\Rightarrow 2x+{60}^o+{60}^o={360}^o$

$\Rightarrow 2x={240}^o$

$\Rightarrow x={120}^o$

$\therefore \angle D={120}^o$

Hence, the answer is ${120}^o.$