Trigonometry & Inverse Trigonometry
Properties of Triangles
Grade 11
Question:
<p>If the angles A, B and C of a triangle are in an arithmetic progression and if a, b and c denote the lengths of the sides opposite to A, B and C respectively, then the value of the expression $\frac{a}{c}\sin 2C + \frac{c}{a}\sin 2A$ is</p>
<p>(a) $\frac{1}{2}$</p>
<p>(b) $\frac{3}{2}$</p>
<p>(c) 1</p>
<p>(d) $\sqrt{3}$</p>
Step-by-Step Solution
Key Concept: Use the arithmetic progression condition on angles to find $B = 60°$, then apply the Sine Rule and Projection Rule to simplify the expression.
<p>Since A, B, C are in AP:</p><p>$2B = A + C$</p><p>Also, $A + B + C = 180°$, so $A + C = 180° - B$</p><p>Therefore, $2B = 180° - B$, which gives $3B = 180°$, so $B = 60°$</p><p>The expression becomes:</p><p>$\frac{a}{c}\sin 2C + \frac{c}{a}\sin 2A = \frac{a}{c}(2\sin C\cos C) + \frac{c}{a}(2\sin A\cos A)$</p><p>$= 2k(a\cos C + c\cos A)$ where $\frac{a}{\sin A} = \frac{c}{\sin C} = k$</p><p>By the Projection Rule: $b = a\cos C + c\cos A$</p><p>Therefore: $= 2kb = 2\sin B = 2\sin 60° = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$</p><p>∴ Answer is (d)</p>
Correct Answer: D