Question
Let $$PQR$$ be a triangle of area $$\Delta $$ 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.
Let $$PQR$$ be a triangle of area $$\Delta $$ 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 $$\frac{{2\sin P - \sin 2P}}{{2\sin P + \sin 2P}}$$ 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 :
$$\eqalign{ & \frac{{2\sin P - 2\sin P\cos P}}{{2\sin P + 2\sin P\cos P}} \cr & = \frac{{1 - \cos P}}{{1 + \cos P}} = \frac{{2{{\sin }^2}\frac{P}{2}}}{{2{{\cos }^2}\frac{P}{2}}} \cr & = {\tan ^2}\frac{P}{2} = \frac{{\left( {s - b} \right)\left( {s - c} \right)}}{{s\left( {s - a} \right)}} = \frac{{{{\left( {\left( {s - b} \right)\left( {s - c} \right)} \right)}^2}}}{{{\Delta ^2}}} \cr & = \frac{{{{\left( {\left( {\frac{1}{2}} \right)\left( {\frac{3}{2}} \right)} \right)}^2}}}{{{\Delta ^2}}} = {\left( {\frac{3}{{4\Delta }}} \right)^2} \cr} $$
$$\eqalign{ & \frac{{2\sin P - 2\sin P\cos P}}{{2\sin P + 2\sin P\cos P}} \cr & = \frac{{1 - \cos P}}{{1 + \cos P}} = \frac{{2{{\sin }^2}\frac{P}{2}}}{{2{{\cos }^2}\frac{P}{2}}} \cr & = {\tan ^2}\frac{P}{2} = \frac{{\left( {s - b} \right)\left( {s - c} \right)}}{{s\left( {s - a} \right)}} = \frac{{{{\left( {\left( {s - b} \right)\left( {s - c} \right)} \right)}^2}}}{{{\Delta ^2}}} \cr & = \frac{{{{\left( {\left( {\frac{1}{2}} \right)\left( {\frac{3}{2}} \right)} \right)}^2}}}{{{\Delta ^2}}} = {\left( {\frac{3}{{4\Delta }}} \right)^2} \cr} $$