[JEE Main 2012] in triangle pqr 3 sin p 4 cos q 6 and 4 sin q
In $\triangle PQR$, $3\sin P+4\cos Q=6$ and $4\sin Q+3\cos P=1$. Then $\angle R$ is
(a) $\frac{5\pi}6$
(b) $\frac\pi6$
(c) $\frac\pi4$
(d) $\frac{3\pi}4$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 4d ago
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Correct answer: (b) $\frac\pi6$
Squaring and adding: $25+24\sin(P+Q)=37\Rightarrow\sin(P+Q)=\frac12$. As $3\sin P+4\cos Q=6$ forces $P$ large, $P+Q=\frac{5\pi}6$, so $R=\frac\pi6$.
JEE Main 2012 · Trigonometry — verified solution by the Vidaara Team.
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