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Question Number 185557 by AROUNAMoussa last updated on 23/Jan/23

Commented by mr W last updated on 23/Jan/23

this is a copy of Q180043

$${this}\:{is}\:{a}\:{copy}\:{of}\:{Q}\mathrm{180043} \\ $$

Commented by Frix last updated on 23/Jan/23

((sin α)/7)=((sin (180−3α))/8)=((sin 2α)/x) 8sin α =7sin 3α 0<α<180 ⇒ sin α=((√(91))/(14)) ⇒ sin 2α =((√(195))/(14)) x=((7sin 2α)/(sin α))=(√(105))

$$\frac{\mathrm{sin}\:\alpha}{\mathrm{7}}=\frac{\mathrm{sin}\:\left(\mathrm{180}−\mathrm{3}\alpha\right)}{\mathrm{8}}=\frac{\mathrm{sin}\:\mathrm{2}\alpha}{{x}} \\ $$$$\mathrm{8sin}\:\alpha\:=\mathrm{7sin}\:\mathrm{3}\alpha \\ $$$$\mathrm{0}<\alpha<\mathrm{180}\:\Rightarrow\:\mathrm{sin}\:\alpha=\frac{\sqrt{\mathrm{91}}}{\mathrm{14}}\:\Rightarrow\:\mathrm{sin}\:\mathrm{2}\alpha\:=\frac{\sqrt{\mathrm{195}}}{\mathrm{14}} \\ $$$${x}=\frac{\mathrm{7sin}\:\mathrm{2}\alpha}{\mathrm{sin}\:\alpha}=\sqrt{\mathrm{105}} \\ $$

Commented by SEKRET last updated on 23/Jan/23

hello sir W . How do you remember them?

$$\:\boldsymbol{\mathrm{hello}}\:\boldsymbol{\mathrm{sir}}\:\:\:\boldsymbol{\mathrm{W}}\:\:.\: \\ $$$$\:\: \\ $$How do you remember them?

Commented by mr W last updated on 23/Jan/23

because i prepaired this question once by myself and solved it.

$${because}\:{i}\:{prepaired}\:{this}\:{question}\:{once} \\ $$$${by}\:{myself}\:{and}\:{solved}\:{it}. \\ $$

Commented by SEKRET last updated on 24/Jan/23

I was wondering why I didn't remember 😁

$$ \\ $$I was wondering why I didn't remember 😁

Answered by HeferH last updated on 23/Jan/23

Commented by HeferH last updated on 23/Jan/23

(7/b) = (8/a) a^2 = 56 − ab (7/b) = (8/a) ⇒ b = ((7a)/8) a^2 = 56 − ((7a^2 )/8) 8a^2 = 448 − 7a^2 a = ((8(√(105)))/(15))⇒ b = (7/8) ∙ ((8(√(105)))/(15)) = ((7(√(105)))/(15)) a + b = ((8(√(105)))/(15))+ ((7(√(105)))/(15)) = ((15(√(105)))/(15)) = (√(105))

$$\:\frac{\mathrm{7}}{{b}}\:=\:\frac{\mathrm{8}}{{a}} \\ $$$$\:{a}^{\mathrm{2}} =\:\mathrm{56}\:−\:{ab} \\ $$$$\:\frac{\mathrm{7}}{{b}}\:=\:\frac{\mathrm{8}}{{a}}\:\Rightarrow\:\:{b}\:=\:\frac{\mathrm{7}{a}}{\mathrm{8}}\: \\ $$$$\:{a}^{\mathrm{2}} \:=\:\mathrm{56}\:−\:\frac{\mathrm{7}{a}^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\:\mathrm{8}{a}^{\mathrm{2}} \:=\:\mathrm{448}\:−\:\mathrm{7}{a}^{\mathrm{2}} \\ $$$$\:{a}\:=\:\frac{\mathrm{8}\sqrt{\mathrm{105}}}{\mathrm{15}}\Rightarrow\:{b}\:=\:\frac{\mathrm{7}}{\mathrm{8}}\:\centerdot\:\frac{\mathrm{8}\sqrt{\mathrm{105}}}{\mathrm{15}}\:=\:\frac{\mathrm{7}\sqrt{\mathrm{105}}}{\mathrm{15}}\: \\ $$$$\:{a}\:+\:{b}\:=\:\:\frac{\mathrm{8}\sqrt{\mathrm{105}}}{\mathrm{15}}+\:\frac{\mathrm{7}\sqrt{\mathrm{105}}}{\mathrm{15}}\:=\:\frac{\mathrm{15}\sqrt{\mathrm{105}}}{\mathrm{15}}\:=\:\sqrt{\mathrm{105}} \\ $$

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