Let PQR be a triangle of area Δ 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 sinP

Engineering

Mathematics

Basic Rules of Properties of Triangle

Question

Let PQR be a triangle of area Δ 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 $\frac{2 sinP - \sin 2 P}{2 sinP + \sin 2 P}$ equals

${(\frac{45}{4 Δ})}^{2}$

$\frac{3}{4 Δ}$

${(\frac{3}{4 Δ})}^{2}$

$\frac{45}{4 Δ}$

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$cosP = \frac{\frac{25}{4} + \frac{49}{4} - 4}{2 (\frac{5}{2}) (\frac{7}{2})} = \frac{74 - 6}{70} = \frac{58}{70} = \frac{29}{35}$

$Now, \frac{2 sinP - \sin 2 P}{2 sinP + \sin 2 P} = \frac{1 - cosP}{1 + cosP}$

$= \frac{1 - \frac{29}{35}}{1 + \frac{29}{35}} = \frac{6}{64} = \frac{3}{32} = {(\frac{3}{4 Δ})}^{2} \Rightarrow (C) is correct$

$As, Δ = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{4 (2) (\frac{3}{2}) (\frac{1}{2})} = \sqrt{6}$

Please subscribe our Youtube channel to unlock this solution.