Question

In a \[\Delta PQR\], if \[\angle QPR={{75}^{o}}\] and \[\angle QRP={{40}^{o}}\], then \[\angle PQR=\]
(a) \[{{65}^{o}}\]
(b) \[{{79}^{o}}\]
(c) \[{{72}^{o}}\]
(d) \[{{89}^{o}}\]

Answer 1

Hint: We know that the sum of the angles of a triangle is equal to \[{{180}^{o}}\] and we have two angles of the triangle, then we will assume the third angle to be a variable and find the sum of all the three angles and equate it to \[{{180}^{o}}\] and we will get the measure of the third angle.

Complete step-by-step answer:

We have been given that in a \[\Delta PQR\], \[\angle QPR={{75}^{o}}\] and \[\angle QRP={{40}^{o}}\]. Let us draw the triangle as shown below. Then, we can mark all the given angles.

Now, let us assume \[\angle PQR\] to be x. As we know that the sum of the angle of a triangle is equal to \[{{180}^{o}}\], we can add all the angles as shown below,

\[\Rightarrow x+{{75}^{o}}+{{40}^{o}}={{180}^{o}}\]

On subtracting \[{{75}^{o}}\] from both the sides of the equality, we get,

\[x+{{75}^{o}}-{{75}^{o}}+{{40}^{o}}={{180}^{o}}-{{75}^{o}}\]

\[\Rightarrow x+40={{105}^{o}}\]

Again on subtracting \[{{40}^{o}}\] from both the sides of the equality, we get,

\[x+{{40}^{o}}-{{40}^{o}}={{105}^{o}}-{{40}^{o}}\]

\[\Rightarrow x={{65}^{o}}\]

Hence, we have found out \[\angle PQR={{65}^{o}}\]

Therefore, the correct answer is the option (a).

Note: Students have to be careful while solving the equation. They need to remember that the sum of the angles of a triangle is equal to \[{{180}^{o}}\]. Students also should not forget to draw the diagram of the given figure, i.e the triangle as it will help you to analyze the problem easily.