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In a $\Delta PQR$ if $3\sin P+4\cos Q=6.4\sin Q+3\cos P=1$ then the angle $R$ is equal to

$\begin{array}{1 1}(a)\;\large\frac{5\pi}{6}&(b)\;\large\frac{\pi}{6}\\(c)\;\large\frac{\pi}{4}&(d)\;\large\frac{3\pi}{4}\end{array}$

1 Answer

Step 1:
Given :
$3\sin P+4\cos Q=6$-----(1)
$4\sin Q+3\cos P=1$-----(2)
Squaring and adding (1) and (2) we get,
$9\sin^2P+16\cos^2Q+2\sin P\cos Q+16\sin^2Q+9\cos^2P+24\sin Q\cos P=36+1$
$\Rightarrow 37$
$9(\sin^2P+\cos^2P)+16(\sin^2Q+\cos^2Q)+24(\sin P\cos Q+\cos P\sin Q)=37$
$\Rightarrow 9+16+24\sin(P+Q)=37$
$\sin^2Q+\cos ^2Q=1$
$\sin A\cos B+\cos A\sin B=\sin(A+B)$
Step 2:
$P+Q=\large\frac{\pi}{6}$ or $\large\frac{5\pi}{6}$
$R=\large\frac{5\pi}{6}$ or $\large\frac{\pi}{6}$
If $R=\large\frac{5\pi}{6}$ then $0 < P < Q <\large\frac{\pi}{6}$
$\Rightarrow \cos Q < 1$ and $\sin P <\large\frac{1}{2}$
$\Rightarrow 3\sin P+4\cos Q < \large\frac{1}{2}$ which is not true.
So $R=\large\frac{\pi}{6}$
answered Oct 18, 2013 by sreemathi.v

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