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In a $\Delta \mathrm{PQR}$, if $3 \sin \mathrm{P}+4 \cos \mathrm{Q}=6$ and $4 \sin \mathrm{Q}+3 \cos \mathrm{P}=1$, then the angle $\mathrm{R}$ is equal to
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The correct answer is:
$\frac{\pi}{6}$
$\frac{\pi}{6}$
$3 \sin P+4 \cos Q=6 \quad\dots(1)$
$4 \sin Q+3 \cos P=1\quad\dots(2)$
From $(1)$ and $(2) \angle P$ is obtuse.
$(3 \sin P+4 \cos Q)^2+(4 \sin Q+3 \cos P)^2=37$
$\Rightarrow 9+16+24(\sin P \cos Q+\cos P \sin Q)=37$
$\Rightarrow 24 \sin (P+Q)=12$
$\Rightarrow \sin (P+Q)=\frac{1}{2} \quad \Rightarrow P+Q=\frac{5 \pi}{6} \quad \Rightarrow R=\frac{\pi}{6}$
$4 \sin Q+3 \cos P=1\quad\dots(2)$
From $(1)$ and $(2) \angle P$ is obtuse.
$(3 \sin P+4 \cos Q)^2+(4 \sin Q+3 \cos P)^2=37$
$\Rightarrow 9+16+24(\sin P \cos Q+\cos P \sin Q)=37$
$\Rightarrow 24 \sin (P+Q)=12$
$\Rightarrow \sin (P+Q)=\frac{1}{2} \quad \Rightarrow P+Q=\frac{5 \pi}{6} \quad \Rightarrow R=\frac{\pi}{6}$
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