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Question Number 153757 by liberty last updated on 10/Sep/21

Commented by mr W last updated on 11/Sep/21

$S_{1} = S_{2} = \frac{85}{4}$
	Answered by mr W last updated on 11/Sep/21

	Commented by mr W last updated on 11/Sep/21
	$((sin (45+α))/(sin α))=((10)/a) (1/( (√2)))((1/(tan α))+1)=((10)/a) ⇒(1/(tan α))=((10(√2))/a)−1 ((sin (45+β))/(sin β))=((10)/b) (1/( (√2)))((1/(tan β))+1)=((10)/b) ⇒(1/(tan β))=((10(√2))/b)−1 α+β=45° tan (α+β)=1 ((tan α+tan β)/(1−tan α tan β))=1 tan α+tan β=1−tan α tan β (1/(tan α))+(1/(tan β))=(1/(tan α))×(1/(tan β))−1 ((10(√2))/a)−1+((10(√2))/b)−1=(((10(√2))/a)−1)×(((10(√2))/b)−1)−1 ((10(√2))/a)+((10(√2))/b)−1=((100)/(ab)) ((10(√2)(a+b))/(ab))−1=((100)/(ab)) ab=15 ⇒a+b=((23(√2))/4) a,b are roots of x^2 −((23(√2))/4)x+15=0 ⇒a,b=(((23±7)(√2))/8)= { ((2(√2))),(((15(√2))/4)) :} (1/(tan α))=((10(√2))/(2(√2)))−1=4 ⇒tan α=(1/4) (1/(tan β))=((10(√2))/((15(√2))/4))=(5/( 3)) ⇒tan β=(3/5) (c/a)=((sin 45)/(sin α)) ⇒c=((2(√2))/( (√2)×(1/( (√(17))))))=2(√(17)) (d/b)=((sin 45)/(sin β)) ⇒d=((15(√2))/( 4(√2)×(3/( (√(34))))))=((5(√(34)))/4) S_1 =((cd sin 45°)/2) =(1/(2(√2)))×2(√(17))×((5(√(34)))/4)=((85)/4) AE=((10)/(cos β))=((10)/(5/( (√(34)))))=2(√(34)) AF=((10)/(cos α))=((10)/( (4/( (√(17))))))=((5(√(17)))/2) S_1 +S_2 =((AE×AF×sin 45°)/2) =(1/(2(√2)))×2(√(34))×((5(√(17)))/( 2))=((85)/2) ⇒S_2 =((85)/2)−((85)/4)=((85)/4)=S_1$
	$\frac{\sin (45 + α)}{\sin α} = \frac{10}{a}$ $\frac{1}{\sqrt{2}} (\frac{1}{\tan α} + 1) = \frac{10}{a}$ $\Rightarrow \frac{1}{\tan α} = \frac{10 \sqrt{2}}{a} - 1$ $\frac{\sin (45 + β)}{\sin β} = \frac{10}{b}$ $\frac{1}{\sqrt{2}} (\frac{1}{\tan β} + 1) = \frac{10}{b}$ $\Rightarrow \frac{1}{\tan β} = \frac{10 \sqrt{2}}{b} - 1$ $α + β = 45 °$ $\tan (α + β) = 1$ $\frac{\tan α + \tan β}{1 - \tan α \tan β} = 1$ $\tan α + \tan β = 1 - \tan α \tan β$ $\frac{1}{\tan α} + \frac{1}{\tan β} = \frac{1}{\tan α} \times \frac{1}{\tan β} - 1$ $\frac{10 \sqrt{2}}{a} - 1 + \frac{10 \sqrt{2}}{b} - 1 = (\frac{10 \sqrt{2}}{a} - 1) \times (\frac{10 \sqrt{2}}{b} - 1) - 1$ $\frac{10 \sqrt{2}}{a} + \frac{10 \sqrt{2}}{b} - 1 = \frac{100}{a b}$ $\frac{10 \sqrt{2} (a + b)}{a b} - 1 = \frac{100}{a b}$ $a b = 15$ $\Rightarrow a + b = \frac{23 \sqrt{2}}{4}$ $a, b a r e r o o t s o f$ $x^{2} - \frac{23 \sqrt{2}}{4} x + 15 = 0$ $\Rightarrow a, b = \frac{(23 \pm 7) \sqrt{2}}{8} = {\begin{cases} 2 \sqrt{2} \\ \frac{15 \sqrt{2}}{4} \end{cases}$ $\frac{1}{\tan α} = \frac{10 \sqrt{2}}{2 \sqrt{2}} - 1 = 4 \Rightarrow \tan α = \frac{1}{4}$ $\frac{1}{\tan β} = \frac{10 \sqrt{2}}{\frac{15 \sqrt{2}}{4}} = \frac{5}{3} \Rightarrow \tan β = \frac{3}{5}$ $\frac{c}{a} = \frac{\sin 45}{\sin α} \Rightarrow c = \frac{2 \sqrt{2}}{\sqrt{2} \times \frac{1}{\sqrt{17}}} = 2 \sqrt{17}$ $\frac{d}{b} = \frac{\sin 45}{\sin β} \Rightarrow d = \frac{15 \sqrt{2}}{4 \sqrt{2} \times \frac{3}{\sqrt{34}}} = \frac{5 \sqrt{34}}{4}$ $S_{1} = \frac{c d \sin 45 °}{2}$ $= \frac{1}{2 \sqrt{2}} \times 2 \sqrt{17} \times \frac{5 \sqrt{34}}{4} = \frac{85}{4}$ $A E = \frac{10}{\cos β} = \frac{10}{\frac{5}{\sqrt{34}}} = 2 \sqrt{34}$ $A F = \frac{10}{\cos α} = \frac{10}{\frac{4}{\sqrt{17}}} = \frac{5 \sqrt{17}}{2}$ $S_{1} + S_{2} = \frac{A E \times A F \times \sin 45 °}{2}$ $= \frac{1}{2 \sqrt{2}} \times 2 \sqrt{34} \times \frac{5 \sqrt{17}}{2} = \frac{85}{2}$ $\Rightarrow S_{2} = \frac{85}{2} - \frac{85}{4} = \frac{85}{4} = S_{1}$
	Commented by Ari last updated on 11/Sep/21
	Mr.W. always, the ratio S2/S1 eshte 1 if dhe angle 0f triangle is 45 grade?
	Commented by mr W last updated on 11/Sep/21

	$y e s, S_{1} = S_{2} i s a l w a y s t r u e .$
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