In a $∆ PQR$ is $3 sinP + 4 cosQ = 6$ and $4 sinQ + 3 cosP = 1$ then the acute angle $R$ is $\dots \dots \dots$

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$\frac{3 π}{4}$

$\frac{π}{6}$

$\frac{5 π}{4}$

$\frac{π}{4}$

answer is B.

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Detailed Solution

In a $∆ PQR$ is $3 sinP + 4 cosQ = 6$ and $4 sinQ + 3 cosP = 1$
On squaring and adding both side and we get,
$\begin{array}{l} (3 \sin P + 4 \cos Q)^{2} + (4 \sin Q + 3 \cos P)^{2} = 36 + 1 \\ \Rightarrow 9 \sin^{2} P + 16 \cos^{2} Q + 24 \sin Pcos Q + 16 \sin^{2} Q + 9 \cos^{2} P + 24 \sin Qcos P = 37 \\ \Rightarrow 9 \sin^{2} P + 9 \cos^{2} P + 16 \sin^{2} Q + 16 \cos^{2} Q + 24 \sin Pcos Q + 24 \sin Qcos P = 37 \\ \Rightarrow 9 (\sin^{2} P + \cos^{2} P) + 16 (\sin^{2} Q + \cos^{2} Q) + 24 (\sin P \cos Q + \sin Q \cos P) = 37 \\ \Rightarrow 9 \times 1 + 16 \times 1 + 24 \sin (P + Q) = 37 \\ \Rightarrow 25 + 24 \sin (P + Q) = 37 \\ \Rightarrow 24 \sin (P + Q) = 37 - 25 \\ \Rightarrow 24 \sin (P + Q) = 12 \\ \Rightarrow \sin (P + Q) = \frac{12}{24} \\ \Rightarrow \sin (P + Q) = \frac{1}{2} \\ \Rightarrow \sin (P + Q) = \sin 30^{\circ} \\ \Rightarrow \sin (P + Q) = \sin 150^{\circ} \\ Now, (P + Q) = 30^{\circ}, 150^{\circ} \\ Now P + Q = 30^{\circ} \\ So, \\ P + Q + R = 180^{\circ} \\ 30^{\circ} + R = 180^{\circ} \\ R = 150^{\circ} = \frac{3 π}{4} \end{array}$