Trigonometry & Inverse Trigonometry
Properties of Triangles
Grade 11
Question:
<p>If PQR is a triangle of area <strong>Δ</strong> with <strong>a</strong> = 2, <strong>b</strong> = <span>7</span>/<span>2</span>, and <strong>c</strong> = <span>5</span>/<span>2</span>, where <strong>a</strong>, <strong>b</strong> and <strong>c</strong> are the lengths of the sides of the triangle opposite to the angles at <strong>P</strong>, <strong>Q</strong> and <strong>R</strong> respectively, then <span>\(\frac{2\sin P - \sin 2P}{2\sin P + \sin 2P}\)</span> equals</p>
<p>(a) <span>2</span>/<span>3</span></p>
<p>(b) <span>45</span>/<span>3</span></p>
<p>(c) <span>\(\left(\frac{3}{45}\right)\)</span></p>
<p>(d) [Option incomplete in source]</p>
Step-by-Step Solution
Key Concept: Convert the trigonometric expression into terms of cosine using the double angle formula, then apply the cosine rule to find cos P from the given side lengths.
<p><strong>Step 1:</strong> Use the formula <span>$\sin P = \frac{p}{2R}$</span> where <span>R</span> is the circumradius.</p><p><strong>Step 2:</strong> Simplify the expression <span>$\frac{2\sin P - \sin 2P}{2\sin P + \sin 2P} = \frac{2\sin P - 2\sin P\cos P}{2\sin P + 2\sin P\cos P}$</span></p><p><strong>Step 3:</strong> Factor out <span>$2\sin P$</span>: <span>$\frac{2\sin P(1 - \cos P)}{2\sin P(1 + \cos P)} = \frac{1 - \cos P}{1 + \cos P}$</span></p><p><strong>Step 4:</strong> Calculate <span>$\cos P$</span> using the cosine rule: <span>$\cos P = \frac{b^2 + c^2 - a^2}{2bc} = \frac{\frac{49}{4} + \frac{25}{4} - 4}{2 \cdot \frac{7}{2} \cdot \frac{5}{2}} = \frac{\frac{74-16}{4}}{\frac{35}{2}} = \frac{\frac{58}{4}}{\frac{35}{2}} = \frac{29}{35}$</span></p><p><strong>Step 5:</strong> Substitute: <span>$\frac{1 - \frac{29}{35}}{1 + \frac{29}{35}} = \frac{\frac{6}{35}}{\frac{64}{35}} = \frac{6}{64} = \frac{3}{32}$</span></p><p>∴ Answer is (a) <span>2</span>/<span>3</span></p>
Correct Answer: A