Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>In a \(\triangle ABC\), \(\tan\dfrac{A}{2} = \dfrac{5}{6}\), \(\tan\dfrac{C}{2} = \dfrac{2}{5}\), then</p>
<p>\(a, c, b\) are in AP.</p>
<p>\(a, b, c\) are in AP.</p>
<p>\(b, a, c\) are in AP.</p>
<p>\(a, b, c\) are in GP.</p>
Step-by-Step Solution
Key Concept: Use the identity tan(A/2)tan(B/2) + tan(B/2)tan(C/2) + tan(C/2)tan(A/2) = 1 (derived from A + B + C = π) to find tan(B/2), then apply half-angle formulas to find the sides or angles.
<p><strong>Step 1:</strong> Since A + B + C = π in a triangle, we have A/2 + B/2 + C/2 = π/2.</p><p>This means B/2 = π/2 - (A/2 + C/2), so tan(B/2) = cot(A/2 + C/2).</p><p><strong>Step 2:</strong> Use the identity for A + B + C = π: tan(A/2)tan(B/2) + tan(B/2)tan(C/2) + tan(C/2)tan(A/2) = 1</p><p>Substituting tan(A/2) = 5/6 and tan(C/2) = 2/5:</p><p>(5/6)tan(B/2) + tan(B/2)(2/5) + (2/5)(5/6) = 1</p><p>tan(B/2)[(5/6) + (2/5)] + (1/3) = 1</p><p>tan(B/2)[(25 + 12)/30] = 2/3</p><p>tan(B/2)(37/30) = 2/3</p><p>tan(B/2) = 20/37</p><p><strong>Step 3:</strong> Alternatively, tan(B/2) = cot(A/2 + C/2) = 1/tan(A/2 + C/2)</p><p>tan(A/2 + C/2) = (5/6 + 2/5)/(1 - (5/6)(2/5)) = (37/30)/(1 - 1/3) = (37/30)/(2/3) = 37/20</p><p>Therefore tan(B/2) = 20/37</p><p>∴ Answer: B</p>
Correct Answer: B