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Question Number 12160 by indreshpatelindresh@435gmail.i last updated on 15/Apr/17
Two angles of a triangle are cot^(−1) 2  and cot^(−1) 3. Then the third angle is
$$\mathrm{Two}\:\mathrm{angles}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{are}\:\mathrm{cot}^{−\mathrm{1}} \mathrm{2} \\ $$$$\mathrm{and}\:\mathrm{cot}^{−\mathrm{1}} \mathrm{3}.\:\mathrm{Then}\:\mathrm{the}\:\mathrm{third}\:\mathrm{angle}\:\mathrm{is} \\ $$
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)=((tg(cot^(−1) 2)+tg(cotg^(−1) 3))/(1−tg(cot^(−1) 2)tg(cotg^(−1) 3)))=  (((1/2)+(1/3))/(1−(1/6)))=((5/6)/(5/6))=1⇒A+B=45⇒C=180−45=135 .■
$${tg}\left({cot}^{−\mathrm{1}} \mathrm{2}+{cot}^{−\mathrm{1}} \mathrm{3}\right)=\frac{{tg}\left({cot}^{−\mathrm{1}} \mathrm{2}\right)+{tg}\left({cotg}^{−\mathrm{1}} \mathrm{3}\right)}{\mathrm{1}−{tg}\left({cot}^{−\mathrm{1}} \mathrm{2}\right){tg}\left({cotg}^{−\mathrm{1}} \mathrm{3}\right)}= \\ $$$$\frac{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{6}}}=\frac{\frac{\mathrm{5}}{\mathrm{6}}}{\frac{\mathrm{5}}{\mathrm{6}}}=\mathrm{1}\Rightarrow{A}+{B}=\mathrm{45}\Rightarrow{C}=\mathrm{180}−\mathrm{45}=\mathrm{135}\:.\blacksquare \\ $$

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