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In a $Δ P Q R$ , if $3 \sin P + 4 \cos Q = 16$ and $4 \sin Q + 3 \cos P = 1$ , Then the angle R is equal to

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π/4

3π/4

5π/6

π/6

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detailed solution

Correct option is D

Squaring and adding the given relations we get16+9+24sin⁡(P+Q)=37⇒sin⁡(P+Q)=1/2⇒sin⁡R=1/2⇒R=π/6 or 5π/6If R=5π/6, then P<π/6⇒3sin⁡P<3/2.⇒3sin⁡P+4cos⁡Q<3/2+4<6So R≠5π/6

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In a ΔPQR, if 3sin⁡P+4cos⁡Q=16 and 4sin⁡Q+3cos⁡P=1, Then the angle R is equal to

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In a $Δ P Q R$ , if $3 \sin P + 4 \cos Q = 16$ and $4 \sin Q + 3 \cos P = 1$ , Then the angle R is equal to