Question

In a parallelogram $ABCD,\angle B$ is ${50}^o$ more than $\angle A$. What is the measure of all the angles of parallelogram, $ABCD$?

Answer 1

We know that sum of two adjacent angles of a parallelogram is ${180}^o$

$\therefore$ $\angle A+\angle B={180}^o$ ------ ( 1 )

It is given that $\angle B$ is ${50}^o$ more than $\angle A$.

$\therefore$ $\angle B=\angle A+{50}^o$ ------ ( 2 )

Now, substituting ( 2 ) in ( 1 ),, we get

$\Rightarrow$ $\angle A+\angle B={180}^o$

$\Rightarrow$ $\angle A+\angle A+{50}^o={180}^o$

$\Rightarrow$ $2\angle A+{50}^o={180}^o$

$\Rightarrow$ $2\angle A={180}^o-{50}^o$

$\Rightarrow$ $2\angle A={130}^o$

$\therefore$ $\angle A={65}^o$

$\Rightarrow$ $\angle B=\angle A+{50}^o={65}^o+{50}^o={115}^o$

In a parallelogram, the opposite angles are equal in magnitude.

$\therefore$ $\angle A=\angle C={65}^o$

$\therefore$ $\angle B=\angle D={115}^o$