In a triangle ABC, if the tangent of half the difference of two angles is equal to one third of the tangent of half the sum of the angles, then the ratio of the sides opposite to the angles is
It is given that: \[ \tan \frac{A-B}{2} = \frac{1}{3} \tan \frac{A+B}{2} = \frac{1}{3} \cot \frac{C}{2}. \] Using Napier's analogy: \[ \tan \frac{A-B}{2} = \frac{a-b}{a+b} \cot \frac{C}{2}. \] From (1), \[ \frac{a-b}{a+b} \cot \frac{C}{2} = \cot \frac{C}{2} \quad \Rightarrow \quad \frac{a-b}{a+b} = 1. \] Thus, \[ a = 2b \quad \text{and the ratio} \quad a:b = 2:1. \]
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