IIT-JEE 2012 · Paper 2 · Mathematics Q42

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MathematicsSingle correct+3 / -1Question 42 of 60

Let $PQR$ be a triangle of area $\Delta$ with $a=2$ , $b=\dfrac72$ and $c=\dfrac52$ , where $a$ , $b$ and $c$ are the lengths of the sides of the triangle opposite to the angles at $P$ , $Q$ and $R$ respectively. Then $\dfrac{2\sin P-\sin2P}{2\sin P+\sin2P}$ equals

\dfrac3{4\Delta}

\dfrac{45}{4\Delta}

\left(\dfrac3{4\Delta}\right)^2

\left(\dfrac{45}{4\Delta}\right)^2

Answer and worked solution

Official answer: C

Simplify.
$\frac{2\sin P-\sin2P}{2\sin P+\sin2P}=\frac{1-\cos P}{1+\cos P}=\tan^2\frac P2.$

Half-angle formula. $s=\frac{2+7/2+5/2}2=4$ , so $s-b=\frac12$ and $s-c=\frac32$ :
$\tan\frac P2=\frac{(s-b)(s-c)}{\Delta}=\frac{3/4}{\Delta}=\frac3{4\Delta}.$

Value. $\left(\dfrac3{4\Delta}\right)^2$ .

(C)

Topics:Properties of triangles Trigonometric ratios and identities

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IIT-JEE 2012 · Paper 2, Mathematics question 42