Question

In parallelogram ABCD if $\angle A=(3x-{20}^{\circ }),\angle B=(y+{15}^{\circ }),\angle C=(x+{40}^{\circ })$, then find the value of x and y .

Answer 1

In a ||gm ABCD,

$\angle A=(3x-{20}^{\circ }),\angle B=y+{15}^{\circ },\angle C=x+{40}^{\circ }$

Now, $\angle A=\angle C$ (Opposite angles of a ||gm)

$\Rightarrow 3x-20=x+{40}^{\circ }$
$\Rightarrow 3x-x={40}^{\circ }+{20}^{\circ }\Rightarrow 2x={60}^{\circ }$
$\Rightarrow x=\frac{{60}^{\circ }}{2}={30}^{\circ }$

and $\angle A+\angle B={180}^{\circ }$ (Sum of the consecutive angles)

$\Rightarrow 3x-{20}^{\circ }+y+{15}^{\circ }={180}^{\circ }$
$\Rightarrow 3x+y-5^{\circ }={180}^{\circ }$
$\Rightarrow 3\times {30}^{\circ }+y-5^{\circ }={180}^{\circ }$
$\Rightarrow {90}^{\circ }-5^{\circ }+y={180}^{\circ }$
$y={180}^{\circ }-{90}^{\circ }+5={95}^{\circ }$
$\therefore x={30}^{\circ },y={95}^{\circ }$