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

Engineering

Mathematics

Trigonometric Ratios of Allied Angles

Question

$\frac{3 π}{4}$

$\frac{π}{4}$

$\frac{π}{6}$

$\frac{5 π}{6}$

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3sin P + 4cos Q = 6 ......(1)

4sin Q + 3cos P = 1 ......(2)

Square and add (1) & (2)

24 sin (P + Q) = 12

$∴ \sin (P + Q) = \frac{1}{2}$

$∴ P + Q = \frac{π}{6} or \frac{5 π}{6}$

But when $P + Q = \frac{π}{6}$ then (1) & (2) not satisfied

$∴ P + Q = \frac{5 π}{6} \Rightarrow R = \frac{π}{6}$

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