If A,B,C are acute angles, $\tan A = \frac{1}{2}, \tan B = \frac{1}{5}, \tan C = \frac{1}{8}$ then $A + B + C =$

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\frac{π}{12}

\frac{π}{4}

\frac{π}{6}

\frac{π}{2}

answer is C.

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Detailed Solution

\tan A = \frac{1}{2}, \tan B = \frac{1}{5}, \tan C = \frac{1}{8}

$\Rightarrow$ $\tan (A + B) = \frac{\frac{1}{2} + \frac{1}{5}}{1 - (\frac{1}{2} . \frac{1}{5})} = \frac{\frac{7}{10}}{\frac{9}{10}} = \frac{7}{9}$

$\Rightarrow$ $\tan (A + B + C) = \frac{\tan (A + B) + \tan C}{1 - \tan (A + B) . \tan C}$

$= \frac{\frac{7}{9} + \frac{1}{8}}{1 - \frac{7}{9} . \frac{1}{8}} = \frac{65}{65} = 1$

$\Rightarrow$ $\tan (A + B + C) = \tan 45$

$\Rightarrow$ $A + B + C = \frac{π}{4}$

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