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

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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

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A

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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.

\(\implies 115^{\circ}+\angle PRQ=180^{\circ}\)

\(\implies \angle PRQ=180^{\circ}-115^{\circ}\)

\(\implies \angle PRQ=65^{\circ}\)

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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