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Question Number 202938 by mr W last updated on 06/Jan/24
Commented by esmaeil last updated on 06/Jan/24
18.4349 ?
$$\mathrm{18}.\mathrm{4349} \\ $$$$? \\ $$$$ \\ $$
Commented by mr W last updated on 06/Jan/24
no.
$${no}. \\ $$
Commented by mr W last updated on 11/Jan/24
Answered by a.lgnaoui last updated on 11/Jan/24
△FBN ∡b+90+60+∡c_1 =180 ∡c_1 =30−∡b BC=q BF=EF=p ((sin b)/q)=((sin (30−b))/p) (1) △AFB ((sin (45+b))/q)=((sina)/p) (2) (1)et (2) ((sin (30−b))/(sin b))=((sina)/(sin (45+b))) a+b=45 ⇒ a=45−b donc ((sin (30−b))/(sin b))=((sin (45−b)/(sin (45+b))) (((1−(√3) tan b))/(2tan b))=((1−tan b)/(1+tan b)) ⇒ (2−(√3) )tan^2 b−(1+(√3))tan b+1=0 ⇒ b=20,81 donc a=24,19 Calcul de x △ AEF ((sin x)/(EF))=((sin a)/(AE)) ((sin x)/p)=((sin a)/(AE)) △ADE ((sin (x+a))/p)=(1/(AE)) ⇒ AE= (p/(sin (x+a))) (( sin x)/p)=((sin a×sin (x+a))/p) sin x=sin a(sin xcos a+cos xsin a) =sin a×cos x(tan xcos a+sin a) tan x=sin a(tan xcos a+sin a) posons t=tan x t=sin acos a×t+sin^2 a t=((sin^2 a)/(1−sin acos a))=0,2681282 Soit x=15°
$$\bigtriangleup\boldsymbol{\mathrm{FBN}}\:\:\measuredangle\mathrm{b}+\mathrm{90}+\mathrm{60}+\measuredangle\mathrm{c}_{\mathrm{1}} =\mathrm{180} \\ $$$$\:\:\:\:\:\:\measuredangle\mathrm{c}_{\mathrm{1}} =\mathrm{30}−\measuredangle\mathrm{b}\:\:\:\mathrm{BC}=\mathrm{q}\:\:\:\:\mathrm{BF}=\mathrm{EF}=\mathrm{p} \\ $$$$\:\:\:\frac{\mathrm{sin}\:\mathrm{b}}{\mathrm{q}}=\frac{\mathrm{sin}\:\left(\mathrm{30}−\mathrm{b}\right)}{\mathrm{p}}\:\:\:\left(\mathrm{1}\right) \\ $$$$\bigtriangleup\boldsymbol{\mathrm{AFB}}\:\:\:\:\frac{\mathrm{sin}\:\left(\mathrm{45}+\mathrm{b}\right)}{\mathrm{q}}=\frac{\mathrm{sina}}{\mathrm{p}}\:\:\:\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\left(\mathrm{1}\right)\mathrm{et}\:\left(\mathrm{2}\right)\:\:\frac{\mathrm{sin}\:\left(\mathrm{30}−\mathrm{b}\right)}{\mathrm{sin}\:\mathrm{b}}=\frac{\mathrm{sina}}{\mathrm{sin}\:\left(\mathrm{45}+\mathrm{b}\right)} \\ $$$$\:\:\:\mathrm{a}+\mathrm{b}=\mathrm{45}\:\:\:\Rightarrow\:\:\:\mathrm{a}=\mathrm{45}−\mathrm{b} \\ $$$$ \\ $$$$\:\:\:\mathrm{donc}\:\:\:\:\:\frac{\mathrm{sin}\:\left(\mathrm{30}−\mathrm{b}\right)}{\mathrm{sin}\:\mathrm{b}}=\frac{\mathrm{sin}\:\left(\mathrm{45}−\mathrm{b}\right.}{\mathrm{sin}\:\left(\mathrm{45}+\mathrm{b}\right)}\:\: \\ $$$$\:\:\:\:\:\:\frac{\left(\mathrm{1}−\sqrt{\mathrm{3}}\:\mathrm{tan}\:\mathrm{b}\right)}{\mathrm{2tan}\:\mathrm{b}}=\frac{\mathrm{1}−\mathrm{tan}\:\mathrm{b}}{\mathrm{1}+\mathrm{tan}\:\mathrm{b}} \\ $$$$\:\:\:\:\:\Rightarrow\:\:\left(\mathrm{2}−\sqrt{\mathrm{3}}\:\right)\mathrm{tan}\:^{\mathrm{2}} \boldsymbol{\mathrm{b}}−\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\mathrm{tan}\:\boldsymbol{\mathrm{b}}+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\Rightarrow\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{b}}=\mathrm{20},\mathrm{81} \\ $$$$\:\:\:\mathrm{donc}\:\:\:\boldsymbol{\mathrm{a}}=\mathrm{24},\mathrm{19} \\ $$$$\:\:\:\mathrm{Calcul}\:\mathrm{de}\:\boldsymbol{\mathrm{x}} \\ $$$$\:\:\bigtriangleup\:\:\mathrm{AEF}\:\:\:\frac{\mathrm{sin}\:\boldsymbol{\mathrm{x}}}{\mathrm{EF}}=\frac{\mathrm{sin}\:\boldsymbol{\mathrm{a}}}{\mathrm{AE}}\:\:\: \\ $$$$\:\:\:\:\:\:\frac{\mathrm{sin}\:\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{p}}}=\frac{\mathrm{sin}\:\boldsymbol{\mathrm{a}}}{\mathrm{AE}} \\ $$$$\:\bigtriangleup\mathrm{ADE}\:\:\:\:\frac{\mathrm{sin}\:\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{a}}\right)}{\boldsymbol{\mathrm{p}}}=\frac{\mathrm{1}}{\mathrm{AE}} \\ $$$$\:\:\:\Rightarrow\:\:\:\:\mathrm{AE}=\:\:\frac{\boldsymbol{\mathrm{p}}}{\mathrm{sin}\:\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{a}}\right)} \\ $$$$\:\:\:\:\frac{\:\:\mathrm{sin}\:\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{p}}}=\frac{\mathrm{sin}\:\boldsymbol{\mathrm{a}}×\mathrm{sin}\:\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{a}}\right)}{\boldsymbol{\mathrm{p}}} \\ $$$$\:\:\:\mathrm{sin}\:\boldsymbol{\mathrm{x}}=\mathrm{sin}\:\boldsymbol{\mathrm{a}}\left(\mathrm{sin}\:\boldsymbol{\mathrm{x}}\mathrm{cos}\:\boldsymbol{\mathrm{a}}+\mathrm{cos}\:\boldsymbol{\mathrm{x}}\mathrm{sin}\:\boldsymbol{\mathrm{a}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{sin}\:\boldsymbol{\mathrm{a}}×\mathrm{cos}\:\boldsymbol{\mathrm{x}}\left(\mathrm{tan}\:\boldsymbol{\mathrm{x}}\mathrm{cos}\:\boldsymbol{\mathrm{a}}+\mathrm{sin}\:\boldsymbol{\mathrm{a}}\right) \\ $$$$\:\:\:\mathrm{tan}\:\boldsymbol{\mathrm{x}}=\mathrm{sin}\:\boldsymbol{\mathrm{a}}\left(\mathrm{tan}\:\boldsymbol{\mathrm{x}}\mathrm{cos}\:\boldsymbol{\mathrm{a}}+\mathrm{sin}\:\boldsymbol{\mathrm{a}}\right) \\ $$$$\:\:\:\:\boldsymbol{\mathrm{posons}}\:\:\:\boldsymbol{\mathrm{t}}=\mathrm{tan}\:\boldsymbol{\mathrm{x}} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{t}}=\mathrm{sin}\:\boldsymbol{\mathrm{a}}\mathrm{cos}\:\boldsymbol{\mathrm{a}}×\boldsymbol{\mathrm{t}}+\mathrm{sin}\:^{\mathrm{2}} \boldsymbol{\mathrm{a}} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{t}}=\frac{\mathrm{sin}\:^{\mathrm{2}} \boldsymbol{\mathrm{a}}}{\mathrm{1}−\mathrm{sin}\:\boldsymbol{\mathrm{a}}\mathrm{cos}\:\boldsymbol{\mathrm{a}}}=\mathrm{0},\mathrm{2681282} \\ $$$$ \\ $$$$\:\:\:\:\boldsymbol{\mathrm{Soit}}\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}=\mathrm{15}°\:\: \\ $$$$\: \\ $$
Commented by mr W last updated on 11/Jan/24
15° is correct.
$$\mathrm{15}°\:{is}\:{correct}. \\ $$
Commented by a.lgnaoui last updated on 11/Jan/24

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