question_answer

Let PQR be a triangle of area \[\Delta \]with \[a=2,b=7/2\] and\[c=5/2\], 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 \[\frac{2\operatorname{sinP}-sin2P}{2\sin P+\sin 2P}\] equals

A) \[\frac{3}{4\Delta }\]

B) \[\frac{45}{4\Delta }\]

C) \[{{\left( \frac{3}{4\Delta } \right)}^{2}}\]

D) \[{{\left( \frac{45}{4\Delta } \right)}^{2}}\]

Correct Answer: C

Solution :

[c] \[\frac{2\sin P-2\sin P\cos P}{2\sin P+2\sin P\cos P}\]

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

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

\[=\frac{{{\left( \left( \frac{1}{2} \right)\left( \frac{3}{2} \right) \right)}^{2}}}{{{\Delta }^{2}}}={{\left( \frac{3}{4\Delta } \right)}^{2}}\]