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

\frac{5 π}{6}

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\frac{π}{6}

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\frac{π}{4}

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\frac{3 π}{4}

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Solution

The correct option is B $\frac{π}{6}$
Ans. $(b)$
$3 sin P + 4 cos Q = 6$
$4 sin Q + 3 cos P = 1$
On squaring and adding we get $sin (P + Q) = \frac{1}{2}$
$\Rightarrow P + Q = \frac{π}{6}$ or $\frac{5 π}{6}$ $\Rightarrow R = \frac{5 π}{6}$ or $\frac{π}{6}$
If $R = \frac{5 π}{6}$ then $0 < P, Q < \frac{π}{6}$
$\Rightarrow cos Q < 1$ and $sin P < \frac{1}{2}$
$\Rightarrow 3 sin P + 4 cos Q < \frac{11}{2}$ Hence $R = \frac{π}{6}$

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