Question Number 193726 by mr W last updated on 18/Jun/23
Answered by som(math1967) last updated on 18/Jun/23
$${a}+{b}+{c} \\ $$$$=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}\:+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}+\mathrm{tan}^{−\mathrm{1}} \mathrm{1} \\ $$$$=\mathrm{tan}^{−\mathrm{1}} \mathrm{1}+\mathrm{45}° \\ $$$$=\mathrm{45}°+\mathrm{45}°=\mathrm{90}° \\ $$
Commented by mr W last updated on 18/Jun/23
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Answered by mr W last updated on 18/Jun/23
Commented by mr W last updated on 18/Jun/23
$${AC}={BC} \\ $$$$\angle{ACB}=\mathrm{90}° \\ $$$$\Rightarrow\angle{ABC}=\mathrm{45}°={a}+{b} \\ $$$${c}=\mathrm{45}° \\ $$$$\Rightarrow{a}+{b}+{c}=\mathrm{45}+\mathrm{45}=\mathrm{90}° \\ $$