Question

Find the measure of each angle of a parallelogram, if one of its angles is ${30}^{\circ }$ less than twice the smallest angle.

Answer 1

ANSWER:

Let ABCD be a parallelogram.
$\therefore \angle A=\angle Cand\angle B=\angle D$
(Opposite angles)

Let $\angle A$ be the smallest angle whose measure is \( x^0 \ )

$\therefore \angle B=(2x-30{)}^0$
Now, $\angle A+\angle B={180}^0$
(Adjacent angles are supplementary)
⇒ $x+2x-{30}^0={180}^0$
⇒ $3x={210}^0$
⇒ $x={70}^0$
$\therefore \angle A=2\times {70}^0-{30}^0={110}^0$
Hence, $\angle A=\angle C={70}^0;\angle B=\angle D={110}^0$