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In the figure side \(QR\) of a \(\Delta PQR\) has been produced to the point \(S\). If \(\angle PRS=115^{\circ }\) and \(\angle P=45^{\circ }\) , then \(\angle Q\) is equal to
( )
A. \(70^{\circ }\)
B. \(105^{\circ }\)
C. \(51^{\circ }\)
D. \(80^{\circ }\)
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Question

In the figure side \(QR\) of a \(\Delta PQR\) has been produced to the point \(S\). If \(\angle PRS=115^{\circ }\) and \(\angle P=45^{\circ }\) , then \(\angle Q\) is equal to
Question: In the figure side QR of a Delta PQR has been produced to the point S. If angle PRS=115^{circ } and angle P=45^{circ } , then angle Q is equal to(   )A. 70^{circ }B. 105^{circ }C. 51^{circ }D. 80^{circ }( )
A. \(70^{\circ }\)
B. \(105^{\circ }\)
C. \(51^{\circ }\)
D. \(80^{\circ }\)

Answer

A
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Solution

Since\(\angle PRS+\angle PRQ=180^{\circ}\)[Linear pair]
\(\implies 115^{\circ}+\angle PRQ=180^{\circ}\)
\(\implies \angle PRQ=180^{\circ}-115^{\circ}\)
\(\implies \angle PRQ=65^{\circ}\)
In \(\triangle PQR\),
\(\angle PRQ+\angle RPQ+\angle PQR=180^{\circ}\)[As the sum of interior angles of a triangle is \(180^{\circ}\) ]
\(\implies \angle PQR=180^{\circ}-65^{\circ}-45^{\circ}\)
\(\implies \angle PQR=70^{\circ}\)
So, \(\angle Q=70 ^{\circ}\)
Thus, option (A) is correct.
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Correct7
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