Question

In the given figure, $l\parallel m$ and $t$ is transversal. If $\angle 1$ and $\angle 2$ are in the ratio $5:7$, find the measure of each of the angles
$\angle 1,\angle 2,\angle 3$ and $\angle 8.$

Answer 1

Given: $l\parallel m$ and $t$ is a transversal.

$\angle 1:\angle 2=5:7$

Let the angles measure $5x$ and $7x.$

$\angle 1+\angle 2={180}^{\circ }$ (linear pair)

$\therefore 5x+7x={180}^{\circ }$

or, $12x={180}^{\circ }$

or, $x={15}^{\circ }$

$\therefore \angle 1=5x=5\left(15\right)={75}^{\circ }$

and $\angle 2=7x=7\left(15\right)={105}^{\circ }$

$\angle 2+\angle 3={180}^{\circ }$ (linear pair)

$\angle 3=180-105={75}^{\circ }$

$\angle 3+\angle 6={180}^{\circ }$ (interior angles on the same side of the transversal are supplementary)

$\angle 6=180-\angle 3={105}^{\circ }$

and $\angle 6=\angle 8={105}^{\circ }$ (vertically opposite angles)

$\therefore \angle 1={75}^{\circ },\angle 2={105}^{\circ },\angle 3={75}^{\circ },\angle 8={105}^{\circ }$