Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

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

Answered by Acem last updated on 07/Dec/22

 x= 90°    “I doupt”   from the common side and the 2 equals bases   ((sin 40)/(sin x))= ((sin 20)/(sin 120−x))  ⇔  ((2 cos 20)/(sin x))= (1/(sin 120−x))   2 cos 20 (((√3)/2) cos x + (1/2) sin x)= sin x   (√3) cos 20 cos x= sin x (1− cos 20)   tan x= (((√3) cos 20)/(1− cos 20)) = 26.988 , x≅ 87.88°

$$\:{x}=\:\cancel{\mathrm{90}°}\:\:\:\:``{I}\:{doupt}'' \\ $$$$\:{from}\:{the}\:{common}\:{side}\:{and}\:{the}\:\mathrm{2}\:{equals}\:{bases} \\ $$$$\:\frac{\mathrm{sin}\:\mathrm{40}}{\mathrm{sin}\:{x}}=\:\frac{\mathrm{sin}\:\mathrm{20}}{\mathrm{sin}\:\mathrm{120}−{x}}\:\:\Leftrightarrow\:\:\frac{\mathrm{2}\:\mathrm{cos}\:\mathrm{20}}{\mathrm{sin}\:{x}}=\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{120}−{x}} \\ $$$$\:\mathrm{2}\:\mathrm{cos}\:\mathrm{20}\:\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{cos}\:{x}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{sin}\:{x}\right)=\:\mathrm{sin}\:{x} \\ $$$$\:\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{20}\:\mathrm{cos}\:{x}=\:\mathrm{sin}\:{x}\:\left(\mathrm{1}−\:\mathrm{cos}\:\mathrm{20}\right) \\ $$$$\:\mathrm{tan}\:{x}=\:\frac{\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{20}}{\mathrm{1}−\:\mathrm{cos}\:\mathrm{20}}\:=\:\mathrm{26}.\mathrm{988}\:,\:{x}\cong\:\mathrm{87}.\mathrm{88}° \\ $$

Commented by HeferH last updated on 07/Dec/22

I would like to see the process :)

$$\left.{I}\:{would}\:{like}\:{to}\:{see}\:{the}\:{process}\::\right) \\ $$$$ \\ $$

Commented by Acem last updated on 07/Dec/22

Haha, i did like you last time :)  in fact by laws it′s very easy, but i would to do  it by Euclidean method. I did, but it was very   complicated, so please give me time to find another   simpler way. Thanks for this question (:

$$\left.{Haha},\:{i}\:{did}\:{like}\:{you}\:{last}\:{time}\::\right) \\ $$$${in}\:{fact}\:{by}\:{laws}\:{it}'{s}\:{very}\:{easy},\:{but}\:{i}\:{would}\:{to}\:{do} \\ $$$${it}\:{by}\:{Euclidean}\:{method}.\:{I}\:{did},\:{but}\:{it}\:{was}\:{very} \\ $$$$\:{complicated},\:{so}\:{please}\:{give}\:{me}\:{time}\:{to}\:{find}\:{another} \\ $$$$\:{simpler}\:{way}.\:{Thanks}\:{for}\:{this}\:{question}\:\left(:\right. \\ $$

Commented by Acem last updated on 07/Dec/22

I doupt that it′s not 90 it′s may be 87.88°

$${I}\:{doupt}\:{that}\:{it}'{s}\:{not}\:\mathrm{90}\:{it}'{s}\:{may}\:{be}\:\mathrm{87}.\mathrm{88}° \\ $$

Commented by Acem last updated on 07/Dec/22

When i solved it with Euclidean method,   “was complicated”i wasn′t know that i had    a mistake in one of assumptions, comparing it   with using laws i got less than 90, that made   me doupt that i have some wrong in formulas   thought it was correct! when i try it again with   Euclidean method i knew that i have a mistake.     However i will not hate that method, sometimes   give me the solution as faster, plus it′s    more fun.

$${When}\:{i}\:{solved}\:{it}\:{with}\:{Euclidean}\:{method}, \\ $$$$\:``{was}\:{complicated}''{i}\:{wasn}'{t}\:{know}\:{that}\:{i}\:{had}\: \\ $$$$\:{a}\:{mistake}\:{in}\:{one}\:{of}\:{assumptions},\:{comparing}\:{it} \\ $$$$\:{with}\:{using}\:{laws}\:{i}\:{got}\:{less}\:{than}\:\mathrm{90},\:{that}\:{made} \\ $$$$\:{me}\:{doupt}\:{that}\:{i}\:{have}\:{some}\:{wrong}\:{in}\:{formulas} \\ $$$$\:{thought}\:{it}\:{was}\:{correct}!\:{when}\:{i}\:{try}\:{it}\:{again}\:{with} \\ $$$$\:{Euclidean}\:{method}\:{i}\:{knew}\:{that}\:{i}\:{have}\:{a}\:{mistake}. \\ $$$$ \\ $$$$\:{However}\:{i}\:{will}\:{not}\:{hate}\:{that}\:{method},\:{sometimes} \\ $$$$\:{give}\:{me}\:{the}\:{solution}\:{as}\:{faster},\:{plus}\:{it}'{s}\: \\ $$$$\:{more}\:{fun}.\: \\ $$

Answered by SEKRET last updated on 04/Jan/23

    (a/(sin(40))) = (y/(sin(x)))        (a/(sin(20))) = (y/(sin(120−x)))    ((sin(40))/(sin(x))) = ((sin(20))/(sin(120−x)))

$$\:\:\:\:\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{sin}}\left(\mathrm{40}\right)}\:=\:\frac{\boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)} \\ $$$$\:\:\:\:\:\:\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{sin}}\left(\mathrm{20}\right)}\:=\:\frac{\boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{sin}}\left(\mathrm{120}−\boldsymbol{\mathrm{x}}\right)} \\ $$$$\:\:\frac{\boldsymbol{\mathrm{sin}}\left(\mathrm{40}\right)}{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)}\:=\:\frac{\boldsymbol{\mathrm{sin}}\left(\mathrm{20}\right)}{\boldsymbol{\mathrm{sin}}\left(\mathrm{120}−\boldsymbol{\mathrm{x}}\right)} \\ $$$$ \\ $$

Commented by Acem last updated on 07/Dec/22

Hi Sir!   can you show me how you solved x

$${Hi}\:{Sir}! \\ $$$$\:{can}\:{you}\:{show}\:{me}\:{how}\:{you}\:{solved}\:{x} \\ $$

Commented by Acem last updated on 07/Dec/22

cause that 2 cos 20 sin 30 ≈ 0.94 ≠ 1

$${cause}\:{that}\:\mathrm{2}\:\mathrm{cos}\:\mathrm{20}\:\mathrm{sin}\:\mathrm{30}\:\approx\:\mathrm{0}.\mathrm{94}\:\neq\:\mathrm{1} \\ $$

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

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

((AB)/(sin 20))=(2/(sin 120))  ⇒AB=((4 sin 20)/( (√3)))=((sin (40+x))/(sin x))  ((4 sin 20)/( (√3)))=((sin 40)/(tan x))+cos 40  cot x=(2/( (√3) cos 20))−(1/(tan 40))  x=cot^(−1) ((2/( (√3) cos 20))−(1/(tan 40)))≈87.878°

$$\frac{{AB}}{\mathrm{sin}\:\mathrm{20}}=\frac{\mathrm{2}}{\mathrm{sin}\:\mathrm{120}} \\ $$$$\Rightarrow{AB}=\frac{\mathrm{4}\:\mathrm{sin}\:\mathrm{20}}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{sin}\:\left(\mathrm{40}+{x}\right)}{\mathrm{sin}\:{x}} \\ $$$$\frac{\mathrm{4}\:\mathrm{sin}\:\mathrm{20}}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{sin}\:\mathrm{40}}{\mathrm{tan}\:{x}}+\mathrm{cos}\:\mathrm{40} \\ $$$$\mathrm{cot}\:{x}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{20}}−\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{40}} \\ $$$${x}=\mathrm{cot}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{20}}−\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{40}}\right)\approx\mathrm{87}.\mathrm{878}° \\ $$

Commented by Acem last updated on 07/Dec/22

Exactly! as i recently got

$${Exactly}!\:{as}\:{i}\:{recently}\:{got} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com