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Question Number 182292 by HeferH last updated on 07/Dec/22

Answered by Acem last updated on 07/Dec/22

$$x=\cancel{90°}‘‘I{doupt}^{″}$$$$fromthecommonsideandthe2equalsbases$$$$\frac{\sin 40}{\sin x}=\frac{\sin 20}{\sin 120-x}\iff \frac{2\cos 20}{\sin x}=\frac{1}{\sin 120-x}$$$$2\cos 20(\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x)=\sin x$$$$\sqrt{3}\cos 20\cos x=\sin x(1-\cos 20)$$$$\tan x=\frac{\sqrt{3}\cos 20}{1-\cos 20}=26.988,x\cong 87.88°$$

Commented by HeferH last updated on 07/Dec/22

$$Iwouldliketoseetheprocess:)$$$$$$

Commented by Acem last updated on 07/Dec/22

$$Haha,ididlikeyoulasttime:)$$$$infactbylaws{it}^{\prime }sveryeasy,butiwouldtodo$$$$itbyEuclideanmethod.Idid,butitwasvery$$$$complicated,sopleasegivemetimetofindanother$$$$simplerway.Thanksforthisquestion(:$$

Commented by Acem last updated on 07/Dec/22

$$Idouptthat{it}^{\prime }snot90{it}^{\prime }smaybe87.88°$$

Commented by Acem last updated on 07/Dec/22

$$WhenisolveditwithEuclideanmethod,$$$$‘‘was{complicated}^{″}i{wasn}^{\prime }tknowthatihad$$$$amistakeinoneofassumptions,comparingit$$$$withusinglawsigotlessthan90,thatmade$$$$medouptthatihavesomewronginformulas$$$$thoughtitwascorrect!whenitryitagainwith$$$$Euclideanmethodiknewthatihaveamistake.$$$$$$$$Howeveriwillnothatethatmethod,sometimes$$$$givemethesolutionasfaster,plus{it}^{\prime }s$$$$morefun.$$

Answered by SEKRET last updated on 04/Jan/23

$$\frac{\mathbf{a}}{\mathbf{sin}\left(40\right)}=\frac{\mathbf{y}}{\mathbf{sin}\left(\mathbf{x}\right)}$$$$\frac{\mathbf{a}}{\mathbf{sin}\left(20\right)}=\frac{\mathbf{y}}{\mathbf{sin}(120-\mathbf{x})}$$$$\frac{\mathbf{sin}\left(40\right)}{\mathbf{sin}\left(\mathbf{x}\right)}=\frac{\mathbf{sin}\left(20\right)}{\mathbf{sin}(120-\mathbf{x})}$$$$$$

Commented by Acem last updated on 07/Dec/22

$$HiSir!$$$$canyoushowmehowyousolvedx$$

Commented by Acem last updated on 07/Dec/22

$$causethat2\cos 20\sin 30\approx 0.94\ne 1$$

Answered by mr W last updated on 07/Dec/22

Commented by mr W last updated on 07/Dec/22

$$\frac{AB}{\sin 20}=\frac{2}{\sin 120}$$$$\Rightarrow AB=\frac{4\sin 20}{\sqrt{3}}=\frac{\sin (40+x)}{\sin x}$$$$\frac{4\sin 20}{\sqrt{3}}=\frac{\sin 40}{\tan x}+\cos 40$$$$\cot x=\frac{2}{\sqrt{3}\cos 20}-\frac{1}{\tan 40}$$$$x={\cot }^{-1}(\frac{2}{\sqrt{3}\cos 20}-\frac{1}{\tan 40})\approx 87.878°$$

Commented by Acem last updated on 07/Dec/22

$$Exactly!asirecentlygot$$

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