Question

In the given diagram, \[PT\] is parallel to \[QR\], size of \[\angle PQR\] is

A. \[{116^0}\]
B. \[{138^0}\]
C. \[{144^0}\]
D. \[{120^0}\]

Answer 1

Hint: First of all, find the value of \[x\] by using the properties of the transversal line between two parallel lines. Then use the angle sum property of a quadrilateral i.e., the sum of the angles is equal to \[{360^0}\] and hence we will obtain the required answer.

Complete step by step solution:
Given \[PT\parallel QR\]
From the figure,
\[
   \Rightarrow \angle PTR = ext.\angle TRQ\,{\text{ }}\left[ {{\text{Alternate angles}}} \right] \\
   \Rightarrow 2x = {128^0} \\
  \therefore x = \dfrac{{{{128}^0}}}{2} = {64^0} \\
\]
In the figure,
\[
   \Rightarrow \angle TRQ + ext.\angle TRQ = {180^0}{\text{ }}\left[ {{\text{pair of interior angles on the same side of the transversal is supplementary}}} \right] \\
   \Rightarrow \angle TRQ + {128^0} = {180^0} \\
   \Rightarrow \angle TRQ = {180^0} - {128^0} \\
  \therefore \angle TRQ = {52^0} \\
\]
We know that sum of the angles in a quadrilateral is equal to \[{360^0}\].
\[
   \Rightarrow \angle PQR + \angle TRQ + \angle QPT + \angle PTR = {360^0} \\
   \Rightarrow \angle PQR + {52^0} + {64^0} + {128^0} = {360^0} \\
   \Rightarrow \angle PQR + {244^0} = {360^0} \\
  \therefore \angle PQR = {360^0} - {244^0} = {116^0} \\
\]
Thus, \[\angle PQR = {116^0}\].

Note: When a transversal intersects two parallel lines the corresponding angles are equal, the vertical opposite angles are equal, the alternate interior angles are equal, the alternate exterior angles are equal and the pair of interior angles on the same side of the transversal is supplementary.