In a Delta PQR,if 3 sin P + 4 cos Q = 6 and 4 | vedclass

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(B) Given equations are:$3 \sin P + 4 \cos Q = 6$  $(1)$$4 \sin Q + 3 \cos P = 1$  $(2)$Squaring and adding $(1)$ and $(2)$:$(3 \sin P + 4 \cos Q)^2 +

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$\frac{\pi}{6}$

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(B) Given equations are:$3 \sin P + 4 \cos Q = 6$  $(1)$$4 \sin Q + 3 \cos P = 1$  $(2)$Squaring and adding $(1)$ and $(2)$:$(3 \sin P + 4 \cos Q)^2 + (4 \sin Q + 3 \cos P)^2 = 6^2 + 1^2$$9 \sin^2 P + 16 \cos^2 Q + 24 \sin P \cos Q + 16 \sin^2 Q + 9 \cos^2 P + 24 \sin Q \cos P = 37$$9(\sin^2 P + \cos^2 P) + 16(\sin^2 Q + \cos^2 Q) + 24(\sin P \cos Q + \cos P \sin Q) = 37$$9(1) + 16(1) + 24 \sin(P + Q) = 37$$25 + 24 \sin(P + Q) = 37$$24 \sin(P + Q) = 12$$\sin(P + Q) = \frac{1}{2}$Since $P + Q + R = \pi$,we have $\sin(P + Q) = \sin(\pi - R) = \sin R$.Thus,$\sin R = \frac{1}{2}$,which implies $R = \frac{\pi}{6}$ or $R = \frac{5\pi}{6}$.If $R = \frac{5\pi}{6}$,then $P + Q = \frac{\pi}{6}$. Since $P, Q > 0$,$\sin P Therefore,$R = \frac{\pi}{6}$.

In a $\Delta PQR$,if $3 \sin P + 4 \cos Q = 6$ and $4 \sin Q + 3 \cos P = 1$,then the angle $R$ is equal to:

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