Trigonometry & Inverse Trigonometry
Properties of Triangles
Grade 11
Question:
<p>If the angles <em>A</em>, <em>B</em> and <em>C</em> of triangle <em>ABC</em> are in arithmetic progression and <em>a</em>, <em>b</em>, <em>c</em> represents length of sides opposite to angles <em>A</em>, <em>B</em> and <em>C</em> respectively, then the value of \(\dfrac{a+c}{\sqrt{(a^2 - ac + c^2)}}\) is:</p>
<p>\(2\cos\dfrac{A+C}{2}\)</p>
<p>\(2\sin\dfrac{A-C}{2}\)</p>
<p>\(2\sin\dfrac{A+C}{2}\)</p>
<p>\(2\cos\dfrac{A-C}{2}\)</p>
Step-by-Step Solution
Key Concept: Since A, B, C are in AP, we have B = 60°. Using the constraint A + B + C = 180° and the sine rule, express the given expression in terms of angles, then substitute B = 60° to simplify.
<p><strong>Step 1:</strong> Since A, B, C are in AP: A + B + C = 180° and B - A = C - B</p><p>This gives: A + C = 2B, so 3B = 180°, thus <strong>B = 60°</strong></p><p><strong>Step 2:</strong> By sine rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$</p><p>Since A + C = 120°, we have C = 120° - A</p><p><strong>Step 3:</strong> Simplify the denominator using the cosine rule:</p><p>$b^2 = a^2 + c^2 - 2ac\cos(60°) = a^2 + c^2 - ac$</p><p>Therefore: $a^2 - ac + c^2 = b^2$</p><p><strong>Step 4:</strong> The expression becomes:</p><p>$\frac{a+c}{\sqrt{b^2}} = \frac{a+c}{b}$</p><p><strong>Step 5:</strong> Using sine rule with B = 60°:</p><p>$\frac{a+c}{b} = \frac{\sin A + \sin C}{\sin 60°} = \frac{\sin A + \sin(120° - A)}{\frac{\sqrt{3}}{2}}$</p><p>Using sum-to-product: $\sin A + \sin(120° - A) = 2\sin(60°)\cos(A - 60°) = \sqrt{3}\cos(A - 60°) \leq \sqrt{3}$</p><p>When this equals √3 (maximum), we get: $\frac{\sqrt{3}}{\sqrt{3}/2} = <strong>2</strong>$</p><p>∴ Answer: <strong>D</strong></p>
Correct Answer: D