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Question Number 12160 by indreshpatelindresh@435gmail.i last updated on 15/Apr/17

$$\mathrm{Two}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{are}{\cot }^{-1}2$$$$\mathrm{and}{\cot }^{-1}3.\mathrm{Then}\mathrm{the}\mathrm{third}\mathrm{angle}\mathrm{is}$$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 15/Apr/17

$$tg({cot}^{-1}2+{cot}^{-1}3)=\frac{tg\left({cot}^{-1}2\right)+tg\left({cotg}^{-1}3\right)}{1-tg\left({cot}^{-1}2\right)tg\left({cotg}^{-1}3\right)}=$$$$\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}=\frac{\frac{5}{6}}{\frac{5}{6}}=1\Rightarrow A+B=45\Rightarrow C=180-45=135.◼$$

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