If in a triangle ABC, $r_{1} = 2 r_{2} = 3 r_{3}$ , then a:b =

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4:5

6:7

2:3

5:4

answer is C.

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Detailed Solution

$\begin{array}{l} r_{1} = 2 r_{2} = 3 r_{3} \Rightarrow s \tan \frac{A}{2} = 2 s \tan \frac{B}{2} = 3 s \tan \frac{C}{2} \\ \Rightarrow \frac{\tan \frac{A}{2}}{6} = \frac{\tan \frac{B}{2}}{3} = \frac{\tan \frac{C}{2}}{2} = k (say) \\ since \frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^{\circ}, we have \\ \tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2} = 1 \\ \Rightarrow (6 k) (3 k) + (3 k) (2 k) + (2 k) (6 k) = 1 \\ \Rightarrow k = \frac{1}{6} \\ ∴ \sin A = \frac{2 \tan \frac{A}{2}}{1 + \tan^{2} \frac{A}{2}} = \frac{12 k}{1 + 36 k^{2}} = 1 \\ and \sin B = \frac{2 \tan \frac{B}{2}}{1 + \tan^{2} \frac{B}{2}} = \frac{6 k}{1 + 9 k^{2}} = \frac{4}{5} \\ Now \frac{a}{b} = \frac{\sin A}{\sin B} = \frac{5}{4} \Rightarrow a : b = 5 : 4 \end{array}$