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Question-159635

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Question Number 159635 by cortano last updated on 19/Nov/21
Commented by cortano last updated on 19/Nov/21
tan α = (x/4) ; tan β = ((3x)/(12(√3))) = (x/(4(√3))) tan β = (1/( (√3))) tan α ⇒α+β+105°=180° ⇒α+β = 75° ⇒tan (α+β)=((1+(1/( (√3))))/(1−(1/( (√3))))) ⇒(((1+(1/( (√3))))tan α)/(1−(1/( (√3)))tan^2 α)) = (((√3)+1)/( (√3)−1)) = 2+(√3) ⇒(1+(√3))tan α = (2+(√3) )((√3)−tan^2 α) ⇒(1+(√3) )tan α =(3+2(√3) )−(2+(√3))tan^2 α ⇒(2+(√3))tan^2 α+(1+(√3) )tan α−(3+2(√3) )=0 ⇒tan α = ((−1−(√3) +(√(4+2(√3)+4(2+(√3))(3+2(√3)))))/(2(2+(√3) ))) ⇒tan α = ((−1−(√3) + (√(4+2(√3) +4(12+7(√3)))))/(2(2+(√3) ))) ⇒tan α =((−1−(√3) +(√(52+30(√3))))/(2(2+(√3)))) =1 ⇒ { ((α =45°)),((β =30°)) :} then x = 4
$$\:\mathrm{tan}\:\alpha\:=\:\frac{{x}}{\mathrm{4}}\:;\:\mathrm{tan}\:\beta\:=\:\frac{\mathrm{3}{x}}{\mathrm{12}\sqrt{\mathrm{3}}}\:=\:\frac{{x}}{\mathrm{4}\sqrt{\mathrm{3}}} \\ $$$$\:\mathrm{tan}\:\beta\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\mathrm{tan}\:\alpha\: \\ $$$$\Rightarrow\alpha+\beta+\mathrm{105}°=\mathrm{180}° \\ $$$$\Rightarrow\alpha+\beta\:=\:\mathrm{75}° \\ $$$$\Rightarrow\mathrm{tan}\:\left(\alpha+\beta\right)=\frac{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\: \\ $$$$\Rightarrow\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\mathrm{tan}\:\alpha}{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mathrm{tan}\:^{\mathrm{2}} \alpha}\:=\:\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\:\sqrt{\mathrm{3}}−\mathrm{1}}\:=\:\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\mathrm{tan}\:\alpha\:=\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)\left(\sqrt{\mathrm{3}}−\mathrm{tan}\:^{\mathrm{2}} \alpha\right) \\ $$$$\Rightarrow\left(\mathrm{1}+\sqrt{\mathrm{3}}\:\right)\mathrm{tan}\:\alpha\:=\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}}\:\right)−\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\mathrm{tan}\:^{\mathrm{2}} \alpha \\ $$$$\Rightarrow\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\mathrm{tan}\:^{\mathrm{2}} \alpha+\left(\mathrm{1}+\sqrt{\mathrm{3}}\:\right)\mathrm{tan}\:\alpha−\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}}\:\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\alpha\:=\:\frac{−\mathrm{1}−\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{4}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}}\right)}}{\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)} \\ $$$$\Rightarrow\mathrm{tan}\:\alpha\:=\:\frac{−\mathrm{1}−\sqrt{\mathrm{3}}\:+\:\sqrt{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\:+\mathrm{4}\left(\mathrm{12}+\mathrm{7}\sqrt{\mathrm{3}}\right)}}{\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)} \\ $$$$\Rightarrow\mathrm{tan}\:\alpha\:=\frac{−\mathrm{1}−\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{52}+\mathrm{30}\sqrt{\mathrm{3}}}}{\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)}\:=\mathrm{1} \\ $$$$\Rightarrow\begin{cases}{\alpha\:=\mathrm{45}°}\\{\beta\:=\mathrm{30}°}\end{cases}\:{then}\:{x}\:=\:\mathrm{4} \\ $$$$ \\ $$

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