Question Number 83030 by Power last updated on 27/Feb/20
Answered by mr W last updated on 27/Feb/20
Commented by mr W last updated on 27/Feb/20
$$\left({a}+{b}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +\left(\mathrm{2}{a}−{b}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{b}=\frac{\mathrm{2}}{\mathrm{3}}{a} \\ $$$$\mathrm{tan}\:\alpha=\frac{{a}}{\mathrm{2}{a}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\beta=\frac{{b}}{\mathrm{2}{a}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\frac{{CB}}{\mathrm{sin}\:\left(\mathrm{45}+\beta\right)}=\frac{{GB}}{\mathrm{sin}\:\beta} \\ $$$${CB}=\mathrm{2}{a}=\frac{\mathrm{3}\:\mathrm{sin}\:\left(\mathrm{45}+\beta\right)}{\mathrm{sin}\:\beta}=\mathrm{3}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}\:\beta}\right) \\ $$$$\Rightarrow\mathrm{2}{a}=\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{3}\right)=\mathrm{6}\sqrt{\mathrm{2}} \\ $$$$\frac{{DH}}{\mathrm{sin}\:\alpha}=\frac{{DC}}{\mathrm{sin}\:\left(\mathrm{45}+\alpha\right)} \\ $$$${x}=\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\left(\mathrm{45}+\alpha\right)}×\mathrm{2}{a}=\frac{\sqrt{\mathrm{2}}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}}×\mathrm{6}\sqrt{\mathrm{2}} \\ $$$${x}=\frac{\sqrt{\mathrm{2}}}{\mathrm{1}+\mathrm{2}}×\mathrm{6}\sqrt{\mathrm{2}}=\mathrm{4} \\ $$
Commented by Power last updated on 27/Feb/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$