Question

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

Answer 1

We know that in parallelogram, sum of adjacent angles is ${180}^o$

$\therefore$ $(3x-4{)}^o+(3x+10{)}^o={180}^o$

$\Rightarrow$ $3x-4+3x+10={180}^o$

$\Rightarrow$ $6x-6={180}^o$

$\Rightarrow$ $6x={174}^o$

$\Rightarrow$ $x=\frac{{174}^o}{6}$

$\Rightarrow$ $x={29}^o.$

$\Rightarrow$ $(3x-4{)}^o=(3\times 29-4{)}^o={83}^o$

$\Rightarrow$ $(3x+10{)}^o=(3\times 29+10{)}^o={97}^o$

We know that, in parallelogram opposite angles are equal.

$\therefore$ Four angles of parallelogram are ${83}^o,{97}^o,{83}^o$ and ${97}^o.$