Question

In the figure, It lines PQ and RS intersect at point T, such that \[\angle PRT\,=\,40{}^\text{o},\,\,\angle RPT\,\,=\,\,95{}^\text{o}\] and \[\angle TQS\,=\,75{}^\text{o}\] find \[\angle SQT.\]

Answer 1

Hint:(i) Use the concept of the angle sum property of the triangle.
(ii) Use the concept of the vertically opposite angles of the triangles.


Complete step by step solution:
Given that
\[\angle PRT\,=\,40{}^\text{o}\]
\[\angle RPT=95{}^\text{o}\]
\[\angle TSQ=75{}^\text{o}\]
In \[\Delta PRT\]
Let \[\angle T=x\]
Use the angle sum property of the triangle
Then
\[\angle RPT+\angle PRT+\angle PTR=180{}^\text{o}\,\,\,\,\,\,\,.....(1)\]
Rewrite the equation \[(1)\]
\[\angle RPT+\angle PRT+\angle PTR=180\]
Substitute the value of \[\angle RPT\] and \[\angle PRT\]
\[95{}^\text{o}+40{}^\text{o}+x=180{}^\text{o}\]
\[x=180{}^\text{o}135{}^\text{o}\]
\[x=45{}^\text{o}\]
Then
\[\angle PTR=45{}^\text{o}\]
\[\angle PTR=\angle QTS\] = (vertically opposite angle)
Then
\[\angle QTS=45{}^\text{o}\]
In \[\Delta QTS\]
Let
\[\angle SQT\,=\,y\]
\[\angle SQT+\angle STQ+\angle QST=180{}^\text{o}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....(2)\]
(Angle sum property of the triangle)
Substitute the value of the \[\angle STQ\,\,and\,\,\angle QST\]in the \[eq(2)\]
\[y+45{}^\text{o}+75{}^\text{o}=180{}^\text{o}\]
Simplify the equation
\[y=180{}^\text{o}-120{}^\text{o}\]
\[y=60{}^\text{o}\]
Then
\[\angle SQT=60{}^\text{o}\]


Note:Students have to be careful about using the concept of the angle sum property of the triangle.
Also, this problem can also be solved by the property of vertically opposite angles.