Question

In figure, AB and CD are parallel lines intersected by a transversal PQ at L and M respectively. If \[\angle LMD{\text{ }} = {\text{ }}35^\circ \] , find \[\angle ALM\] and \[\angle PLA\].

Answer 1

Hint: Here we will have to apply concepts of –
Sum of all the angles on a straight line is \[180^\circ \] & Alternate interior angles are equal when two lines are parallel & there is a transversal. Applying this concept of geometry, we can get values of angles asked for in the above question.

Complete step-by-step answer:
Given:
AB and CD are parallel lines (AB II CD) intersected by a transversal PQ at L and M respectively.
[ Transversal – A line that intersects two lines on the same plane intersecting at two different points.]
\[\angle LMD{\text{ }} = {\text{ }}35^\circ \]
To find : \[\angle ALM\] & \[\angle PLA\]
In this figure, \[\angle ALM\] and $\angle LMD$ are anterior interior angles.
[ Anterior internal angles are a pair of angles on the inner side of each side of each of those two lines but on the opposite sides of the transversal]
We know that anterior interior angles are equal when two lines are parallel & there is a transversal.
AB and CD are parallel lines, AB II CD & PQ is transversal.
\[{
  \therefore \;\angle ALM = \angle LMD = 35^\circ \\
    \\
} \]
Hence, the value of \[\angle ALM\] is \[35^\circ \].
Now to find value of \[\angle PLA\],
We need to know that, Sum of all the angles on a straight line is \[180^\circ \] .
In the above figure, on line segment LM,
\[\angle ALM\]+ \[\angle PLA\] = \[180^\circ \]
$ \Rightarrow \angle PLA = 180^\circ - \angle ALM$
$ = 180^\circ - 35^\circ $
\[ = 145^\circ \]
$\therefore \angle PLA = 145^\circ $
Hence the value of \[\angle PLA\] is $145^\circ $

Note: Alternate interior angles are equal, when lines are parallel to each other & there is a transversal so by implementing this concept we can get a linear equation to be solved carefully to get the ultimate answer.