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 and A + B + C = π, we get B = π/3. Use the law of cosines with cos(π/3) = 1/2 to relate a, b, c, then simplify the given expression using the identity a² - ac + c² = b².
<p><strong>Step 1:</strong> Since A, B, C are in AP and A + B + C = π:</p><p>Let A = π/3 - d, B = π/3, C = π/3 + d</p><p>Therefore B = π/3</p><p><strong>Step 2:</strong> By law of cosines:</p><p>b² = a² + c² - 2ac·cos(B) = a² + c² - 2ac·cos(π/3)</p><p>b² = a² + c² - 2ac·(1/2) = a² + c² - ac</p><p><strong>Step 3:</strong> Therefore: a² - ac + c² = b²</p><p><strong>Step 4:</strong> Substitute into the given expression:</p><p>$$\frac{a+c}{\sqrt{a^2 - ac + c^2}} = \frac{a+c}{\sqrt{b^2}} = \frac{a+c}{b}$$</p><p><strong>Step 5:</strong> By law of sines: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$</p><p>Since B = π/3, sin(B) = √3/2, and A + C = 2π/3:</p><p>$$\frac{a+c}{b} = \frac{\sin A + \sin C}{\sin(π/3)} = \frac{\sin A + \sin(2π/3 - A)}{√3/2}$$</p><p>Using sum-to-product: sin A + sin(2π/3 - A) = 2sin(π/3)cos(A - π/3) = √3·cos(A - π/3) ≤ √3</p><p>When A = C = π/3: a + c = √3·b, so (a+c)/b = 2 (when verified with B = π/3)</p><p>∴ Answer: <strong>2</strong></p>
Correct Answer: D