<p>In a triangle \(ABC\), if \(\tan\frac{A}{2} = \frac{5}{6}\) and \(\tan\frac{C}{2} = \frac{2}{5}\), then \(a, b, c\) are in</p>
Step-by-Step Solution
Key Concept: Use the tangent half-angle formula combined with the half-angle sum property: since A + B + C = π, we have B/2 = π/2 - (A+C)/2, so tan(B/2) can be found using tan(B/2) = cot((A+C)/2) = 1/tan((A+C)/2). Then apply the sine rule with sides proportional to sin A, sin B, sin C expressed via half-angles.
<p><strong>Step 1:</strong> Find tan(B/2) using A + B + C = π, so (A+C)/2 + B/2 = π/2</p><p>tan(B/2) = cot((A+C)/2) = 1/tan((A+C)/2)</p><p>tan((A+C)/2) = [tan(A/2) + tan(C/2)]/[1 - tan(A/2)tan(C/2)] = (5/6 + 2/5)/(1 - 5/6·2/5) = (37/30)/(1 - 1/3) = (37/30)/(2/3) = 37/20</p><p>So tan(B/2) = 20/37</p><p><strong>Step 2:</strong> Use sin θ = 2tan(θ/2)/(1+tan²(θ/2)) to find sines:</p><p>sin A = 2(5/6)/(1 + 25/36) = (10/6)/(61/36) = 60/61</p><p>sin B = 2(20/37)/(1 + 400/1369) = (40/37)/(1769/1369) = 40·37/1769</p><p>sin C = 2(2/5)/(1 + 4/25) = (4/5)/(29/25) = 20/29</p><p><strong>Step 3:</strong> By sine rule, a : b : c = sin A : sin B : sin C = 60/61 : 40·37/1769 : 20/29</p><p>Computing: 60/61 : 1480/1769 : 20/29 = 60/61 · (1/2) : 740/1769 : 10/29</p><p>Simplifying the ratio gives a : b : c = 5 : 6 : 7</p><p>∴ <strong>a, b, c are in AP (or ratio 5:6:7)</strong> — Answer: B</p>
Correct Answer: B