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Question Number 159635 by cortano last updated on 19/Nov/21

Commented by cortano last updated on 19/Nov/21
$tan α = (x/4) ; tan β = ((3x)/(12(√3))) = (x/(4(√3))) tan β = (1/( (√3))) tan α ⇒α+β+105°=180° ⇒α+β = 75° ⇒tan (α+β)=((1+(1/( (√3))))/(1−(1/( (√3))))) ⇒(((1+(1/( (√3))))tan α)/(1−(1/( (√3)))tan^2 α)) = (((√3)+1)/( (√3)−1)) = 2+(√3) ⇒(1+(√3))tan α = (2+(√3) )((√3)−tan^2 α) ⇒(1+(√3) )tan α =(3+2(√3) )−(2+(√3))tan^2 α ⇒(2+(√3))tan^2 α+(1+(√3) )tan α−(3+2(√3) )=0 ⇒tan α = ((−1−(√3) +(√(4+2(√3)+4(2+(√3))(3+2(√3)))))/(2(2+(√3) ))) ⇒tan α = ((−1−(√3) + (√(4+2(√3) +4(12+7(√3)))))/(2(2+(√3) ))) ⇒tan α =((−1−(√3) +(√(52+30(√3))))/(2(2+(√3)))) =1 ⇒ { ((α =45°)),((β =30°)) :} then x = 4$
$\tan α = \frac{x}{4}; \tan β = \frac{3 x}{12 \sqrt{3}} = \frac{x}{4 \sqrt{3}}$ $\tan β = \frac{1}{\sqrt{3}} \tan α$ $\Rightarrow α + β + 105 ° = 180 °$ $\Rightarrow α + β = 75 °$ $\Rightarrow \tan (α + β) = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}}$ $\Rightarrow \frac{(1 + \frac{1}{\sqrt{3}}) \tan α}{1 - \frac{1}{\sqrt{3}} \tan^{2} α} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = 2 + \sqrt{3}$ $\Rightarrow (1 + \sqrt{3}) \tan α = (2 + \sqrt{3}) (\sqrt{3} - \tan^{2} α)$ $\Rightarrow (1 + \sqrt{3}) \tan α = (3 + 2 \sqrt{3}) - (2 + \sqrt{3}) \tan^{2} α$ $\Rightarrow (2 + \sqrt{3}) \tan^{2} α + (1 + \sqrt{3}) \tan α - (3 + 2 \sqrt{3}) = 0$ $\Rightarrow \tan α = \frac{- 1 - \sqrt{3} + \sqrt{4 + 2 \sqrt{3} + 4 (2 + \sqrt{3}) (3 + 2 \sqrt{3})}}{2 (2 + \sqrt{3})}$ $\Rightarrow \tan α = \frac{- 1 - \sqrt{3} + \sqrt{4 + 2 \sqrt{3} + 4 (12 + 7 \sqrt{3})}}{2 (2 + \sqrt{3})}$ $\Rightarrow \tan α = \frac{- 1 - \sqrt{3} + \sqrt{52 + 30 \sqrt{3}}}{2 (2 + \sqrt{3})} = 1$ $\Rightarrow {\begin{cases} α = 45 ° \\ β = 30 ° \end{cases} t h e n x = 4$
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