Question
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:
A.
$$\frac{{5\pi }}{6}$$
B.
$$\frac{{\pi }}{6}$$
C.
$$\frac{{\pi }}{4}$$
D.
$$\frac{{3\pi }}{4}$$
Answer :
$$\frac{{\pi }}{6}$$
Solution :
Given $$3\sin P + 4\cos Q = 6\,\,\,\,.....\left( {\text{i}} \right)$$
$$4\sin Q + 3\cos P = 1\,\,\,\,.....\left( {{\text{ii}}} \right)$$
Squaring and adding (i) & (ii) we get
$$\eqalign{
& 9{\sin ^2}P + 16{\cos ^2}Q + 24\sin P\cos Q + 16{\sin ^2}Q + 9{\cos ^2}P + 24\sin Q\cos P = 36 + 1 = 37 \cr
& \Rightarrow \,\,9\left( {{{\sin }^2}P + {{\cos }^2}P} \right) + 16\left( {{{\sin }^2}Q + {{\cos }^2}Q} \right) + 24\left( {\sin P\cos Q + \cos P\sin Q} \right) = 37 \cr
& \Rightarrow \,\,9 + 16 + 24\sin \left( {P + Q} \right) = 37 \cr
& \left[ {\because \,\,{{\sin }^2}\theta + {{\cos }^2}\theta = 1\,\,{\text{and }}\sin A\cos B + \cos A\sin B = \sin \left( {A + B} \right)} \right] \cr
& \Rightarrow \,\,\sin \left( {P + Q} \right) = \frac{1}{2} \cr
& \Rightarrow \,\,P + Q = \frac{\pi }{6}\,\,{\text{or }}\frac{{5\pi }}{6} \cr
& \Rightarrow \,\,R = \frac{{5\pi }}{6}\,\,{\text{or }}\frac{\pi }{6}\,\,\left( {\because \,\,P + Q + R = \pi } \right) \cr
& {\text{If }}R = \frac{{5\pi }}{6}\,\,{\text{then 0}} < P,Q < \frac{\pi }{6} \cr
& \Rightarrow \,\,\cos Q < 1\,\,{\text{and }}\sin P < \frac{1}{2} \cr} $$
$$ \Rightarrow \,\,3\sin P + 4\cos Q < \frac{{11}}{2}$$ which is not true.
$${\text{So, }}R = \frac{\pi }{6}$$