Question

In figure, $AB$ and $CD$ are parallel lines intersected by a transversal $PQ$ at $L$ and $M$ respectively. If $\angle CMQ={60}^o$, find all other angles in the figure.

Answer 1

As per the diagram

We know that when a line is intersected by a transversal then

Corresponding angles are equal

Alternate interior angles are equal

Vertically opposite angles are equal

and also Sum of all the angles in a straight line is ${180}^o$

Given,

$\angle CMQ={60}^o$

=>$\angle LMD={60}^o$ [ ∵ vertically opposite angles are equal]

=>$\angle QMD=180-{60}^o$ [ ∵sum of angles in a straight line is ${180}^o$]

$\therefore$ $\angle QMD={120}^o$

=>$\angle CML={120}^o$ [ ∵ vertically opposite angles are equal]

=>$\angle MLB={120}^o$ [ ∵ Alternate interior angles are equal]

=>$\angle LMD=MLA={60}^o$ [ ∵ Alternate interior angles are equal]

$\therefore$ $\angle MLA=PLB={60}^o$ [ ∵ vertically opposite angles are equal]

=>$\angle MLB=\angle ALP={120}^o$ [ ∵ vertically opposite angles are equal]

$\therefore$ $\angle CMQ={60}^o$, $\angle QMD={120}^o$, $\angle CML={120}^o$, $\angle LMD={60}^o$,

$\angle MLA={60}^o$ , $\angle MLB={120}^o$, $\angle ALP={120}^o$ and $\angle PLB={60}^o$ .