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Question Number 182078 by Acem last updated on 04/Dec/22

Answered by mr W last updated on 05/Dec/22

((sin 30)/(sin (30−θ)))=((sin (30+θ))/(sin θ)) ((sin θ)/2)=(((cos θ+(√3) sin θ)(cos θ−(√3) sin θ))/4) 2 sin θ=cos^2 θ−3 sin^2 θ 4 sin^2 θ+2 sin θ−1=0 sin θ=((−1+(√5))/4) ⇒θ=18°

$$\frac{\mathrm{sin}\:\mathrm{30}}{\mathrm{sin}\:\left(\mathrm{30}−\theta\right)}=\frac{\mathrm{sin}\:\left(\mathrm{30}+\theta\right)}{\mathrm{sin}\:\theta} \\ $$$$\frac{\mathrm{sin}\:\theta}{\mathrm{2}}=\frac{\left(\mathrm{cos}\:\theta+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta\right)\left(\mathrm{cos}\:\theta−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta\right)}{\mathrm{4}} \\ $$$$\mathrm{2}\:\mathrm{sin}\:\theta=\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:\theta \\ $$$$\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}\:\mathrm{sin}\:\theta−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{sin}\:\theta=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\:\Rightarrow\theta=\mathrm{18}° \\ $$

Commented by manxsol last updated on 04/Dec/22

Commented by manxsol last updated on 04/Dec/22

thans,Sir W

$${thans},{Sir}\:{W} \\ $$

Commented by mr W last updated on 05/Dec/22

thanks for adding the explanation!

$${thanks}\:{for}\:{adding}\:{the}\:{explanation}! \\ $$

Commented by manxsol last updated on 05/Dec/22

Sometimes I don′t understand a part of your explanation. That′s why I′m looking for how?he did it. You an Sir Acem are a great guide

$${Sometimes}\:{I}\:{don}'{t}\:{understand}\: \\ $$$${a}\:{part}\:{of}\:{your}\:{explanation}.\: \\ $$$${That}'{s}\:{why}\:\:{I}'{m}\:{looking}\: \\ $$$${for}\:{how}?{he}\:{did}\:{it}. \\ $$$${You}\:\:{an}\:{Sir}\:{Acem}\:{are}\:{a}\:{great}\:{guide} \\ $$

Commented by Acem last updated on 05/Dec/22

Very well Sir W, thank you for your sharing! P.s, just fix sin θ= (((√5) −1)/4) Thank you

$$\:{Very}\:{well}\:{Sir}\:{W},\:{thank}\:{you}\:{for}\:{your}\:{sharing}! \\ $$$$\:{P}.{s},\:{just}\:{fix}\:\mathrm{sin}\:\theta=\:\frac{\sqrt{\mathrm{5}}\:−\mathrm{1}}{\mathrm{4}}\:\:\:{Thank}\:{you} \\ $$$$ \\ $$

Commented by Acem last updated on 05/Dec/22

@Mr.Manxsol, Thank you with all my heart! And thank you again for sharing :) Well, It′s a good news that you today have solved our friend Mr. W ′s code (: Go ahead! By the way, what′s about the black board you write on? i liked it, is it real or a plasma screen? My bro, you can check out the solution with other method below, but keep your use of the laws cause sometimes we may fail at the Euclidian method, but it′s more interesting! Hope you′re always doing well!

$$\:@{Mr}.{Manxsol},\:{Thank}\:{you}\:{with}\:{all}\:{my}\:{heart}! \\ $$$$\left.\:{And}\:{thank}\:{you}\:{again}\:{for}\:{sharing}\::\right) \\ $$$$ \\ $$$$\:{Well},\:{It}'{s}\:{a}\:{good}\:{news}\:{that}\:{you}\:{today}\:{have}\:{solved} \\ $$$$\:\:{our}\:{friend}\:{Mr}.\:{W}\:'{s}\:{code}\:\left(:\:{Go}\:{ahead}!\right. \\ $$$$ \\ $$$$\:{By}\:{the}\:{way},\:{what}'{s}\:{about}\:{the}\:{black}\:{board}\:{you} \\ $$$$\:{write}\:{on}?\:\:{i}\:{liked}\:{it},\:{is}\:{it}\:{real}\:{or}\:{a}\:{plasma}\:{screen}? \\ $$$$ \\ $$$$\:{My}\:{bro},\:{you}\:{can}\:{check}\:{out}\:{the}\:{solution}\:{with} \\ $$$$\:{other}\:{method}\:{below},\:{but}\:{keep}\:{your}\:{use}\:{of}\:{the}\:{laws} \\ $$$$\:{cause}\:{sometimes}\:{we}\:{may}\:{fail}\:{at}\:{the} \\ $$$$\:{Euclidian}\:{method},\:{but}\:{it}'{s}\:{more}\:{interesting}! \\ $$$$ \\ $$$$\:{Hope}\:{you}'{re}\:{always}\:{doing}\:{well}! \\ $$$$ \\ $$

Answered by Acem last updated on 05/Dec/22

Euclidean Solution

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Euclidean}\:{Solution} \\ $$

Commented by Acem last updated on 05/Dec/22

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