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Question-185557




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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