Question Number 140138 by henderson last updated on 04/May/21
Answered by mr W last updated on 04/May/21
Commented by mr W last updated on 04/May/21
$${yellow}\:{angle}\:{x}=\alpha+\beta+\mathrm{90}° \\ $$
Commented by mr W last updated on 04/May/21
$${Method}\:{I}: \\ $$
Commented by mr W last updated on 04/May/21
Commented by mr W last updated on 04/May/21
$${it}\:{can}\:{be}\:{easily}\:{proved}\:{that}\:{the} \\ $$$${red}\:{triangle}\:{is}\:{an}\:{isosceles}\:{right}− \\ $$$${angled}\:{triangle},\:{therefore} \\ $$$$\alpha+\beta=\mathrm{45}° \\ $$$$\Rightarrow{x}=\alpha+\beta+\mathrm{90}°=\mathrm{135}° \\ $$
Commented by mr W last updated on 04/May/21
$${Method}\:{II}: \\ $$$$\mathrm{tan}\:\alpha=\frac{{a}−{b}}{{a}+{b}} \\ $$$$\mathrm{tan}\:\beta=\frac{{b}}{{a}} \\ $$$$\mathrm{tan}\:\left(\alpha+\beta\right)=\frac{\frac{{a}−{b}}{{a}+{b}}+\frac{{b}}{{a}}}{\mathrm{1}−\frac{{a}−{b}}{{a}+{b}}×\frac{{b}}{{a}}} \\ $$$$\:\:\:\:\:\:=\frac{{a}^{\mathrm{2}} −{ab}+{ab}+{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{ab}−{ab}+{b}^{\mathrm{2}} }=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow\alpha+\beta=\mathrm{45}° \\ $$$$\Rightarrow{x}=\alpha+\beta+\mathrm{90}°=\mathrm{135}° \\ $$