Question

In a quadrilateral $ABCD$, $AB\parallel CD$ . If $\angle A:\angle D=2:3$ and $\angle B:\angle C=7:8$, find the measure of each angle.

Answer 1

$\angle A:\angle D=2:3$

Let $\angle A$ and $\angle D$ be $2x$ and $3x.$

$\Rightarrow$ $\angle B:\angle C=7:8$

Let $\angle A$ and $\angle D$ be $7y$ and $8y.$

Since, $AB\parallel CD$ and $AD$ is a transversal.

$\therefore$ $\angle A+\angle D={180}^o.$ [ Interior angles on same side of transversal ]

$\Rightarrow$ $2x+3x={180}^o.$

$\Rightarrow$ $5x={180}^o.$

$\Rightarrow$ $x={36}^o$

$\Rightarrow$ $\angle A=2\times {36}^o={72}^o$

$\Rightarrow$ $\angle D=3\times {36}^o={108}^o.$

Since, $AB\parallel CD$ and $BC$ is a transversal.

$\therefore$ $\angle B+\angle C={180}^o.$ [ Interior angles on same side of transversal ]

$\Rightarrow$ $7y+8y={180}^o$

$\Rightarrow$ $15y={180}^o$

$\therefore$ $y={12}^o$

$\Rightarrow$ $\angle B=7\times {12}^o={84}^o$

$\Rightarrow$ $\angle C=8\times {12}^o={96}^o$