Question

If PQR is a triangle of area $\triangle$ 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 a t P ,Q and R, respectively. Then, $ \frac{ 2 \, sin \, P - sin \, 2 P }{ 2 \, sin \, P + sin \, 2P }$ equals

• A. $ \frac{ 3}{4 \triangle} $
• B. $ \frac{ 45}{4 \triangle}$
• C. $\bigg( \frac{ 3}{4 \triangle} \bigg)^2$
• D. $\bigg(\frac{ 45}{4 \triangle}\bigg)^5$

Answer 1

The Correct Option is C

PLAN If $\triangle$ ABC has sides a, b, c Then, tan (A/2) = $ \sqrt{ \frac{ (s - b) \, (s - a)}{ s \, (s - a)}} $ where, s = $ \frac{a + b + c}{ 2} \Rightarrow s = \frac{ 2 + \frac{ 7}{2} + \frac{5}{2} }{2} = 4 $ $ \therefore \frac{ 2 \, sin \, P - sin \, 2 P }{ 2 \, sin \, P + sin \, 2P } = \frac{ 2 sin \, P (1 - cos \, P )}{ 2 \, sin \, P (1 + cos \, P )}$ $\hspace26mm$ = $ \frac{ 2 \, sin^2 (P/2)}{2 \, cos^2 (P/2)} = tan^2 \, (P/2)$ $\Rightarrow \frac{ (s - b) \, (s - c)}{ s \, (s - a)} \times \frac{ (s- b) \, (s - c)}{ (s - b) \, (s - c)}$ = $ \frac{ [ (s - b)^2 \, (s - c)^2 ] }{ \triangle^2 } = \frac{\bigg( 4 - \frac{7}{2}\bigg)^2 \, \bigg( 4 - \frac{5}{2}\bigg)^2 }{ \triangle^2 } = \bigg( \frac{ 3}{ 4 \triangle } \bigg)^2 $