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

\frac{3 π}{4}

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\frac{5 π}{6}

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\frac{π}{6}

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\frac{π}{4}

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Solution

The correct option is C $\frac{5 π}{6}$
Given trignometric equations are:
$3 sin P + 4 cos Q = 6$ -------(1)
$4 sin Q + 3 cos P = 1$ -------(2)

Squaring both equations (1) and (2) and adding them, we get
$\Rightarrow 9 ({sin}^{2} P + {cos}^{2} P) + 16 ({sin}^{2} Q + {cos}^{2} Q) + 24 (sin Q cos P + cos Q sin P) = 36 + 1$

$\Rightarrow 9 + 16 + 24 sin (P + Q) = 37$

$∴ sin (P + Q) = \frac{37 - 25}{24} = \frac{12}{24} = \frac{1}{2}$

$∴ P + Q = 30 °$

Hence, angle $R = 180 ° - 30 ° = 150 ° = \frac{5 π}{6}$ radian

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