Question-154088

Question Number 154088 by iloveisrael last updated on 14/Sep/21

Commented by Tawa11 last updated on 14/Sep/21

$$\mathrm{Weldone}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 14/Sep/21

tan 3α=2 tan 2α ((tan α(3−tan^2 α))/(1−3 tan^2 α))=2×((2 tan α)/(1−tan^2 α)) ((3−tan^2 α)/(1−3 tan^2 α))=(4/(1−tan^2 α)) tan^4 α+8 tan^2 α−1=0 ⇒tan^2 α=(√(17))−4 ⇒tan α=(√((√(17))−4)) tan 2α=((2(√((√(17))−4)))/(1−((√(17))−4)))=((2(√((√(17))−4)))/(5−(√(17)))) x=BE×tan 2α=12×sin 2α×tan 2α =12×((2(√((√(17))−4)))/(5−(√(17))))×((2(√((√(17))−4)))/( (√(26−6(√(17)))))) =((24)/( 4)) =6

$$\mathrm{tan}\:\mathrm{3}\alpha=\mathrm{2}\:\mathrm{tan}\:\mathrm{2}\alpha \\ $$$$\frac{\mathrm{tan}\:\alpha\left(\mathrm{3}−\mathrm{tan}^{\mathrm{2}} \:\alpha\right)}{\mathrm{1}−\mathrm{3}\:\mathrm{tan}^{\mathrm{2}} \:\alpha}=\mathrm{2}×\frac{\mathrm{2}\:\mathrm{tan}\:\alpha}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\alpha} \\ $$$$\frac{\mathrm{3}−\mathrm{tan}^{\mathrm{2}} \:\alpha}{\mathrm{1}−\mathrm{3}\:\mathrm{tan}^{\mathrm{2}} \:\alpha}=\frac{\mathrm{4}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\alpha} \\ $$$$\mathrm{tan}^{\mathrm{4}} \:\alpha+\mathrm{8}\:\mathrm{tan}^{\mathrm{2}} \:\alpha−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}^{\mathrm{2}} \:\alpha=\sqrt{\mathrm{17}}−\mathrm{4} \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\sqrt{\sqrt{\mathrm{17}}−\mathrm{4}} \\ $$$$\mathrm{tan}\:\mathrm{2}\alpha=\frac{\mathrm{2}\sqrt{\sqrt{\mathrm{17}}−\mathrm{4}}}{\mathrm{1}−\left(\sqrt{\mathrm{17}}−\mathrm{4}\right)}=\frac{\mathrm{2}\sqrt{\sqrt{\mathrm{17}}−\mathrm{4}}}{\mathrm{5}−\sqrt{\mathrm{17}}} \\ $$$${x}={BE}×\mathrm{tan}\:\mathrm{2}\alpha=\mathrm{12}×\mathrm{sin}\:\mathrm{2}\alpha×\mathrm{tan}\:\mathrm{2}\alpha \\ $$$$=\mathrm{12}×\frac{\mathrm{2}\sqrt{\sqrt{\mathrm{17}}−\mathrm{4}}}{\mathrm{5}−\sqrt{\mathrm{17}}}×\frac{\mathrm{2}\sqrt{\sqrt{\mathrm{17}}−\mathrm{4}}}{\:\sqrt{\mathrm{26}−\mathrm{6}\sqrt{\mathrm{17}}}} \\ $$$$=\frac{\mathrm{24}}{\:\mathrm{4}} \\ $$$$=\mathrm{6} \\ $$

Commented by mr W last updated on 14/Sep/21