JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Properties of Triangle

Let

P QR

be a triangle of area

△

with

a = 2, b = \frac{7}{2}

and

c = \frac{5}{2}

, 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

\frac{2 s i n P - s i n 2 P}{2 s i n P + s i n 2 P}

equals

Answer: (C)

Solution

Simplify the expression:

\frac{2 sin P - 2 sin P cos P}{2 sin P + 2 sin P cos P} = \frac{1 - cos P}{1 + cos P} = \frac{2 sin ^{2} \frac{P}{2}}{2 cos ^{2} \frac{P}{2}} = tan^{2} \frac{P}{2}

Using the half-angle formula:

tan^{2} \frac{P}{2} = \frac{( s - b ) ( s - c )}{s ( s - a )}

, where

s = \frac{a + b + c}{2} = \frac{6}{2} = 3

.

So

(s - b) = \frac{1}{2}

(s - c) = \frac{3}{2}

(s - a) = 1

. Thus:

tan^{2} \frac{P}{2} = \frac{( s - b ) ( s - c )}{s ( s - a )} = \frac{\frac{1}{2} \cdot \frac{3}{2}}{3 \cdot 1} = \frac{3/4}{3} = \frac{1}{4}

But we also have

△ = s (s - a) (s - b) (s - c) = 3 \cdot 1 \cdot \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{2}

.

Therefore:

tan^{2} \frac{P}{2} = \frac{(( s - b ) ( s - c ) ) ^{2}}{△ ^{2}} = \frac{( \frac{3}{4} ) ^{2}}{△ ^{2}} = (\frac{3}{4△})^{2}

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Single correct option