Question

In a parallelogram $ABCD$, if $\angle A={(2x+5)}^o$ and $\angle B={(3x-5)}^o$, find the value of $x$ and the measure of each angle of the parallelogram.

Answer 1

We know that the opposite angles are equal in a parallelogram

Consider parallelogram $ABCD$

So we get

$\angle A=\angle C={(2x-25)}^o$

$\angle B=\angle D={(3x-5)}^o$

We know that the sum of all the angles of a parallelogram is ${360}^o$

so it can be written as

$\angle A+\angle B+\angle C+\angle D={360}^o$

By subtituting the values in the above equation

$(2x+25)+(3x-5)+(2x+25)+(3x-5)={360}^o$

by addition we get

$10x+{40}^o={360}^o$

$10x={320}^o$

$x={32}^o$

now substituting the value $x$

$\angle A=\angle C={(2x+25)}^o={\left(2\right(32)+25)}^o$

$\angle A=\angle C={(54+25)}^o$

by addition

$\angle A=\angle C={89}^o$

$\angle B=\angle D={(3x-5)}^o={\left(3\right(32)-5)}^o$

$\angle B=\angle D={(96-5)}^o$

by subtraction

$\angle B=\angle D={91}^o$

therefore $x={32}^o$, $\angle A=\angle C={89}^o$ and $\angle B=\angle D={91}^o$