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$I n a Δ P Q R, i f 3 \sin P + 4 \cos Q = 6 a n d 4 \sin Q + 3 \cos P = 1$ then the acute angle $R$ is equal to

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Correct option is A

We have 3sinP+4cosQ=6, 4sinQ+3cosP=1. Squaring on both sides and adding, we get 9sin2P+cos2P+16sin2Q+cos2Q+24sinPcosQ+cosPsinQ=37 ⇒24sinP+Q=37-25=12 ⇒sinπ-R=12 ∵P+Q+R=π ⇒sinR=12 ⇒R=π6∵R is acute

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In a ΔPQR, if 3sinP+4cosQ=6 and 4sinQ+3cosP=1then the acute angle Ris equal to

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$I n a Δ P Q R, i f 3 \sin P + 4 \cos Q = 6 a n d 4 \sin Q + 3 \cos P = 1$ then the acute angle $R$ is equal to