Question
Let $$PQR$$ be a triangle of area $$\Delta $$ with $$a = 2,$$ $$b = \frac{7}{2}\,{\text{and }}c = \frac{5}{2};$$ where $$a, b,$$ and $$c$$ are the lengths of the sides of the triangle opposite to the angles at $$PQ$$ and $$R$$ respectively. Then $$\frac{{2\sin P - \sin 2P}}{{2\sin P + \sin 2P}}$$ equals.
A.
$$\frac{3}{{4\Delta }}$$
B.
$$\frac{45}{{4\Delta }}$$
C.
$${\left( {\frac{3}{{4\Delta }}} \right)^2}$$
D.
$${\left( {\frac{45}{{4\Delta }}} \right)^2}$$
Answer :
$${\left( {\frac{3}{{4\Delta }}} \right)^2}$$
Solution :
We have,
$$\eqalign{
& \frac{{2\sin P - \sin 2P}}{{2\sin P + \sin 2P}} \cr
& = \frac{{2\sin P - 2\sin P\cos P}}{{2\sin P + 2\sin P\cos P}} \cr
& = \frac{{1 - \cos P}}{{1 - \sin P}} \cr
& = \frac{{2{{\sin }^2}\frac{P}{2}}}{{2{{\cos }^2}\frac{P}{2}}} \cr
& = {\tan ^2}\frac{P}{2} \cr
& = \frac{{\left( {s - b} \right)\left( {s - c} \right)}}{{s\left( {s - a} \right)}} \cr
& {\text{where }}s = \frac{{a + b + c}}{2} \cr
& = \frac{{{{\left( {s - b} \right)}^2}{{\left( {s - c} \right)}^2}}}{{s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}} \cr
& = \frac{{{{\left( {a + c - b} \right)}^2}{{\left( {a + b - c} \right)}^2}}}{{16.{\Delta ^2}}} \cr
& = \frac{{1 \times 9}}{{16{\Delta ^2}}} \cr
& = {\left( {\frac{3}{{4\Delta }}} \right)^2} \cr} $$