Question Number 194903 by cortano12 last updated on 19/Jul/23
Commented by cortano12 last updated on 19/Jul/23
$$\:\:\:{x}\:=\:\sqrt{\mathrm{39}}\:\:\left(×\right) \\ $$$$\:\:\:{x}\:=\:\sqrt{\mathrm{30}}\:\left(\checkmark\right) \\ $$
Answered by MM42 last updated on 19/Jul/23
$${tan}\left(\alpha+\theta\right)=\frac{\mathrm{10}}{{x}} \\ $$$$\frac{{tan}\alpha+{tan}\theta}{\mathrm{1}−{tan}\alpha×{tan}\theta}=\frac{\mathrm{10}}{{x}} \\ $$$$\frac{\frac{\mathrm{3}}{{x}}+{tan}\theta}{\mathrm{1}−\frac{\mathrm{3}}{{x}}{tan}\theta}\:=\frac{\mathrm{10}}{{x}}\:\Rightarrow\:\mathrm{3}{x}+{x}^{\mathrm{2}} {tan}\theta=\mathrm{10}{x}−\mathrm{30}{tan}\theta \\ $$$$\Rightarrow{tan}\theta=\frac{\mathrm{7}{x}}{{x}^{\mathrm{2}} +\mathrm{30}}\Rightarrow\theta={tan}^{−\mathrm{1}} \left(\frac{\mathrm{7}{x}}{{x}^{\mathrm{2}} +\mathrm{30}}\right) \\ $$$$\Rightarrow\theta^{'} \:=\:\frac{\frac{\mathrm{7}\left({x}^{\mathrm{2}} +\mathrm{30}\right)−\mathrm{2}{x}\left(\mathrm{7}{x}\right)}{\left({x}^{\mathrm{2}} +\mathrm{30}\right)^{\mathrm{2}} }}{\mathrm{1}+\left({x}^{\mathrm{2}} +\mathrm{30}\right)^{\mathrm{2}} }\:=\mathrm{0} \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{30}}\:\:\checkmark\: \\ $$$$ \\ $$
Commented by MM42 last updated on 19/Jul/23
