AIEEE 2012 (Offline) 29th April Code AQuestion 1
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 :
Trigonometric Ratio & Identities
Solution
Concept. Squaring and adding the two equations brings out $\sin(P+Q)$. Formulas used. (1) $\sin P\cos Q+\cos P\sin Q=\sin(P+Q)$(2) In a triangle, $R=\pi-(P+Q)$Step 1 - Square and add. $$9(\sin^2P+\cos^2P)+16(\sin^2Q+\cos^2Q)+24(\sin P\cos Q+\cos P\sin Q)=36+1.$$ By (1), $25+24\sin(P+Q)=37$, so $\sin(P+Q)=\frac12$. Step 2 - Two cases. $P+Q=\frac\pi6$ or $\frac{5\pi}6$. Step 3 - Reject $\frac\pi6$. Then $P<\frac\pi6$, so $3\sin P<\frac32$, and $4\cos Q\le4$. The first equation would give less than $5.5$, not $6$. Step 4 - Value. $P+Q=\frac{5\pi}6$, so by (2) $R=\frac\pi6$. Answer: (A)