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Question Number 177432 by HeferH last updated on 05/Oct/22

Commented by Ar Brandon last updated on 05/Oct/22

30° ?

$$\mathrm{30}°\:? \\ $$

Commented by HeferH last updated on 05/Oct/22

I′m not sure

$${I}'{m}\:{not}\:{sure} \\ $$$$\: \\ $$

Commented by Ar Brandon last updated on 05/Oct/22

Commented by HeferH last updated on 05/Oct/22

Commented by HeferH last updated on 05/Oct/22

x + α + 140°=180 x = α + 20° x = 30°

$$\:{x}\:+\:\alpha\:\:+\:\mathrm{140}°=\mathrm{180} \\ $$$$\:{x}\:=\:\alpha\:+\:\mathrm{20}° \\ $$$$\:{x}\:=\:\mathrm{30}° \\ $$$$\: \\ $$

Commented by Tawa11 last updated on 05/Oct/22

Great sirs

$$\mathrm{Great}\:\mathrm{sirs} \\ $$

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

how can you say α=10°? where is AE=CD?

$${how}\:{can}\:{you}\:{say}\:\alpha=\mathrm{10}°?\:{where}\:{is} \\ $$$${AE}={CD}? \\ $$

Commented by HeferH last updated on 05/Oct/22

It′s one of those magic lines

$$\:{It}'{s}\:{one}\:{of}\:{those}\:{magic}\:{lines}\: \\ $$$$\: \\ $$

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

what′s the proof for its correctness?

$${what}'{s}\:{the}\:{proof}\:{for}\:{its}\:{correctness}? \\ $$

Commented by HeferH last updated on 05/Oct/22

I know it′s an insecure way, but I see so many teachers draw certain lines that at first are valid but then I think “That′s not possible, is it?”, in this case I′m trusting that by forming 140° with the drawn segment the end terminates at a point F on AC such that AF = FD = DC Probably I had luck

$$\:{I}\:{know}\:{it}'{s}\:{an}\:{insecure}\:{way},\:{but}\:{I}\:{see}\:{so} \\ $$$$\:{many}\:{teachers}\:{draw}\:{certain}\:{lines}\:{that} \\ $$$$\:{at}\:{first}\:{are}\:{valid}\:{but}\:{then}\:{I}\:{think}\:``{That}'{s} \\ $$$$\:{not}\:{possible},\:{is}\:{it}?'',\:{in}\:{this}\:{case}\:{I}'{m} \\ $$$$\:{trusting}\:{that}\:{by}\:{forming}\:\mathrm{140}°\:{with}\:{the}\: \\ $$$$\:{drawn}\:{segment}\:{the}\:{end}\:{terminates}\:{at} \\ $$$$\:{a}\:{point}\:{F}\:{on}\:{AC}\:{such}\:{that}\:{AF}\:=\:{FD}\:=\:{DC} \\ $$$$\: \\ $$$$\:{Probably}\:{I}\:{had}\:{luck} \\ $$

Answered by Ar Brandon last updated on 05/Oct/22

(i) ((EB)/(sin40°))=(h/(sin80°)) ⇒EB=h((sin40°)/(sin80°))=(h/(2cos40°)) (ii) x+EB=h ⇒x=(1−(1/(2cos40°)))h=(((2cos40°−1)/(2cos40°)))h (iii) ((CE)/(sin60°))=(h/(sin80°)) ⇒CE=h((sin60°)/(sin80°)) (iv) x+DE=CE ⇒DE=CE−x=(((sin60°)/(sin80°))−((2cos40°−1)/(2cos40°)))h Applying cosine rule in △ADE AD=(√(x^2 +DE^2 −2x.DE.cos100°)) =h(√((((2cos40°−1)/(2cos40°)))^2 +(((sin60°)/(sin80°))−((2cos40°−1)/(2cos40°)))^2 −2(((2cos40°−1)/(2cos40°)))(((sin60°)/(sin80°))−((2cos40°−1)/(2cos40°)))cos100°)) ≈h(√(0,347296355^2 +0,532088886^2 −2(0,347296355)(0,532088886)cos100°)) ≈h(√(0,467911113))=0,684040286h Applying Sine rule we have ((sinϑ)/x)=((sin100°)/(AD)) ⇒ϑ=sin^(−1) ((x/(AD))sin100°)=sin^(−1) ((h/(AD))∙((2cos40°−1)/(2cos40°))∙sin100°) ϑ=sin^(−1) ((((2cos40°−1)(sin100°))/( (0,684040286)(2cos40°))))=30°

$$\left({i}\right)\:\frac{\mathrm{EB}}{\mathrm{sin40}°}=\frac{{h}}{\mathrm{sin80}°}\:\Rightarrow\mathrm{EB}={h}\frac{\mathrm{sin40}°}{\mathrm{sin80}°}=\frac{{h}}{\mathrm{2cos40}°} \\ $$$$\left({ii}\right)\:{x}+\mathrm{EB}={h}\:\Rightarrow{x}=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2cos40}°}\right){h}=\left(\frac{\mathrm{2cos40}°−\mathrm{1}}{\mathrm{2cos40}°}\right){h} \\ $$$$\left({iii}\right)\:\frac{\mathrm{CE}}{\mathrm{sin60}°}=\frac{{h}}{\mathrm{sin80}°}\:\Rightarrow\mathrm{CE}={h}\frac{\mathrm{sin60}°}{\mathrm{sin80}°} \\ $$$$\left({i}\mathrm{v}\right)\:{x}+\mathrm{DE}=\mathrm{CE}\:\Rightarrow\mathrm{DE}=\mathrm{CE}−{x}=\left(\frac{\mathrm{sin60}°}{\mathrm{sin80}°}−\frac{\mathrm{2cos40}°−\mathrm{1}}{\mathrm{2cos40}°}\right){h} \\ $$$$\mathrm{Applying}\:\mathrm{cosine}\:\mathrm{rule}\:\mathrm{in}\:\bigtriangleup\mathrm{ADE} \\ $$$$\mathrm{AD}=\sqrt{{x}^{\mathrm{2}} +\mathrm{DE}^{\mathrm{2}} −\mathrm{2}{x}.\mathrm{DE}.\mathrm{cos100}°} \\ $$$$\:\:\:\:\:\:\:\:={h}\sqrt{\left(\frac{\mathrm{2cos40}°−\mathrm{1}}{\mathrm{2cos40}°}\right)^{\mathrm{2}} +\left(\frac{\mathrm{sin60}°}{\mathrm{sin80}°}−\frac{\mathrm{2cos40}°−\mathrm{1}}{\mathrm{2cos40}°}\right)^{\mathrm{2}} −\mathrm{2}\left(\frac{\mathrm{2cos40}°−\mathrm{1}}{\mathrm{2cos40}°}\right)\left(\frac{\mathrm{sin60}°}{\mathrm{sin80}°}−\frac{\mathrm{2cos40}°−\mathrm{1}}{\mathrm{2cos40}°}\right)\mathrm{cos100}°} \\ $$$$\:\:\:\:\:\:\:\:\approx{h}\sqrt{\mathrm{0},\mathrm{347296355}^{\mathrm{2}} +\mathrm{0},\mathrm{532088886}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{0},\mathrm{347296355}\right)\left(\mathrm{0},\mathrm{532088886}\right)\mathrm{cos100}°} \\ $$$$\:\:\:\:\:\:\:\:\approx{h}\sqrt{\mathrm{0},\mathrm{467911113}}=\mathrm{0},\mathrm{684040286}{h} \\ $$$$\mathrm{Applying}\:\mathrm{Sine}\:\mathrm{rule}\:\mathrm{we}\:\mathrm{have} \\ $$$$\frac{\mathrm{sin}\vartheta}{{x}}=\frac{\mathrm{sin100}°}{\mathrm{AD}}\:\Rightarrow\vartheta=\mathrm{sin}^{−\mathrm{1}} \left(\frac{{x}}{\mathrm{AD}}\mathrm{sin100}°\right)=\mathrm{sin}^{−\mathrm{1}} \left(\frac{{h}}{\mathrm{AD}}\centerdot\frac{\mathrm{2cos40}°−\mathrm{1}}{\mathrm{2cos40}°}\centerdot\mathrm{sin100}°\right) \\ $$$$\vartheta=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\left(\mathrm{2cos40}°−\mathrm{1}\right)\left(\mathrm{sin100}°\right)}{\:\left(\mathrm{0},\mathrm{684040286}\right)\left(\mathrm{2cos40}°\right)}\right)=\mathrm{30}° \\ $$

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

((AD)/(sin 100))=((AE)/(sin x)) ...(i) ((sin 20)/(AD))=((sin (x−20))/(CD)) ...(ii) (i)×(ii): ((sin 20)/(sin 100))=((sin (x−20))/(sin x)) (1/(cos 10))=(1/(tan 20))−(1/(tan x)) tan x=(1/((1/(tan 20))−(1/(cos 10)))) ⇒x=tan^(−1) ((1/((1/(tan 20))−(1/(cos 10)))))=30°

$$\frac{{AD}}{\mathrm{sin}\:\mathrm{100}}=\frac{{AE}}{\mathrm{sin}\:{x}}\:\:\:\:...\left({i}\right) \\ $$$$\frac{\mathrm{sin}\:\mathrm{20}}{{AD}}=\frac{\mathrm{sin}\:\left({x}−\mathrm{20}\right)}{{CD}}\:\:\:...\left({ii}\right) \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$$\frac{\mathrm{sin}\:\mathrm{20}}{\mathrm{sin}\:\mathrm{100}}=\frac{\mathrm{sin}\:\left({x}−\mathrm{20}\right)}{\mathrm{sin}\:{x}} \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{10}}=\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{20}}−\frac{\mathrm{1}}{\mathrm{tan}\:{x}} \\ $$$$\mathrm{tan}\:{x}=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{20}}−\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{10}}} \\ $$$$\Rightarrow{x}=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{20}}−\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{10}}}\right)=\mathrm{30}° \\ $$

Commented by Ar Brandon last updated on 05/Oct/22

Nice Sir !

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