Question

In the figure if $l||m$ and $\angle 1=(2x+y{)}^{\circ }$ , $\angle 4=(x+2y{)}^{\circ }$ and $\angle 6=(3y+20{)}^{\circ }$. Find $\angle 7$ and $\angle 8$.

Answer 1

Given Line $l$ and $m$ are paralle and line $n$ is transversal line.
Thus, $\angle 4=\angle 6$ as they are alternate interior angles.
$\Rightarrow x+2y=3y+20$
$\Rightarrow x=y+20$ .....(1)
Now,$\angle 1+\angle 4={180}^{\circ }$ (angle of straight line)
$\Rightarrow 2x+y+x+2y={180}^{\circ }$
$\Rightarrow 3x+3y={180}^{\circ }$
$\Rightarrow x+y={60}^{\circ }$
Substitute equation(i) in above equation, we get
$\Rightarrow y+20+y=60$
$\Rightarrow 2y=60-20$
$\Rightarrow 2y={40}^{\circ }$
$\therefore y={20}^{\circ }$
Thus, $\angle 6=3y+20$
$=3\times 20+20$
$=60+20$
$={80}^{\circ }$
Now, $\angle 6+\angle 7={180}^{\circ }$ (angle of straight line)
$\Rightarrow 80+\angle 7=180$
$\Rightarrow \angle 7=180-80$
$\therefore \angle 7={100}^{\circ }$
And $\angle 8=\angle 6={80}^{\circ }$ (vertical opposite angles)
Hence, $\angle 7={100}^{\circ }$ and $\angle 8={80}^{\circ }$