Question

Two adjacent angles of a parallelogram are $(3x-4{)}^{\circ }$ and $(3x+10{)}^{\circ }$. Find the angles of the parallelogram.

• A. $\angle A={83}^{\circ },\angle B={97}^{\circ },\angle C={83}^{\circ },\angle D={97}^{\circ }$
• B. $\angle A={97}^{\circ },\angle B={83}^{\circ },\angle C={97}^{\circ },\angle D={83}^{\circ }$
• C. $\angle A={78}^{\circ },\angle B={89}^{\circ },\angle C={78}^{\circ },\angle D={89}^{\circ }$
• D. $\angle A={83}^{\circ },\angle B={83}^{\circ },\angle C={97}^{\circ },\angle D={97}^{\circ }$

Answer 1

The correct option is A $\angle A={83}^{\circ },\angle B={97}^{\circ },\angle C={83}^{\circ },\angle D={97}^{\circ }$


Let in parallelogram ABCD,
$\angle A=3x-4^{\circ }and\angle B=3x+{10}^{\circ }$

AD || BC and AB is a transversal.
Then, $\angle A+\angle B={180}^{\circ }$
(Interior angles of same side of the transversal is supplementary)

$\Rightarrow (3x-4{)}^{\circ }+(3x+10{)}^{\circ }={180}^{\circ }\Rightarrow 6x+6^{\circ }={180}^{\circ }\Rightarrow 6x={180}^{\circ }-6^{\circ }={174}^{\circ }\therefore x=\frac{{174}^{\circ }}{6}={29}^{\circ }\therefore \angle A=3x-4=3\times {29}^{\circ }-4^{\circ }={87}^{\circ }-4^{\circ }={83}^{\circ }$
$\angle B=(3x+10{)}^{\circ }=3\times {29}^{\circ }+{10}^{\circ }={87}^{\circ }+{10}^{\circ }={97}^{\circ }$

But $\angle C=\angle A$ and $\angle D=\angle B$
(Opposite angles of a parallelogram are equal)
$\therefore \angle C={83}^{\circ }and\angle D={97}^{\circ }$
Hence, $\angle A={83}^{\circ },\angle B={97}^{\circ },\angle C={83}^{\circ }$ and $\angle D={97}^{\circ }.$