Question Number 30160 by naka3546 last updated on 17/Feb/18
Commented by ajfour last updated on 18/Feb/18
$${is}\:{it}\:{right}\:? \\ $$
Commented by ajfour last updated on 17/Feb/18
Commented by ajfour last updated on 17/Feb/18
![∠C = 180°−α−2β = 180°−42°−72° = 66° ∠A = 𝛗 = 180°−2α−∠C ⇒𝛗 =180°−84°−66° = 30° let x= θ and EM be ⊥ on AC Then tan θ = ((EM)/(AM)) = ((EM)/(AD+DM)) =((c sin 𝛃)/(acos 𝛗−bcos 2𝛃+ccos 𝛃)) From △ABD (a/(sin (α+φ)))=(b/(sin φ)) and From △BDE (c/(sin α)) = (b/(sin (α+β))) tan 𝛉 = ((sin β)/((a/c)cos φ−(b/c)cos 2β+cos β)) =((sin β)/(((b sin (α+φ)cos φ)/(c sin φ))−(b/c)cos 2𝛃+cos 𝛃)) =((sin 𝛃)/(((sin (α+β))/(sin α))[((sin (α+φ)cos φ)/(sin φ))−cos 2β]+cos β)) α+φ = 42°+30° =72° = 2β , so =((sin 𝛃)/(((sin (α+β))/(sin α))[((sin (2β−φ))/(sin φ))]+cos β)) 2β−φ = α =((sin β)/(((sin (α+β))/(sin φ))+cos β)) ⇒ tan θ = ((sin 36°)/(2sin 78°+cos 36°)) =((sin 12°(3−4sin^2 12°))/(2cos 12°+cos 12°(4cos^2 12°−3))) = tan 12° [((3−4sin^2 12°)/(2+4−4sin^2 12°−3))] tan 𝛉 = tan 12° ⇒ x = 𝛉 = 12° .](Q30170.png)
$$\angle{C}\:=\:\mathrm{180}°−\alpha−\mathrm{2}\beta \\ $$$$\:\:\:\:\:\:\:\:=\:\mathrm{180}°−\mathrm{42}°−\mathrm{72}°\:=\:\mathrm{66}° \\ $$$$\angle{A}\:=\:\boldsymbol{\phi}\:=\:\mathrm{180}°−\mathrm{2}\alpha−\angle{C} \\ $$$$\:\:\:\:\:\:\:\:\:\Rightarrow\boldsymbol{\phi}\:=\mathrm{180}°−\mathrm{84}°−\mathrm{66}°\:=\:\mathrm{30}° \\ $$$${let}\:\:\:\:{x}=\:\theta\:\:{and}\:{EM}\:{be}\:\bot\:{on}\:{AC} \\ $$$${Then} \\ $$$$\:\:\:\mathrm{tan}\:\theta\:=\:\frac{{EM}}{{AM}}\:=\:\frac{{EM}}{{AD}+{DM}} \\ $$$$\:\:\:\:=\frac{\boldsymbol{{c}}\:\mathrm{sin}\:\boldsymbol{\beta}}{\boldsymbol{{a}}\mathrm{cos}\:\boldsymbol{\phi}−\boldsymbol{{b}}\mathrm{cos}\:\mathrm{2}\boldsymbol{\beta}+\boldsymbol{{c}}\mathrm{cos}\:\boldsymbol{\beta}} \\ $$$${From}\:\bigtriangleup{ABD} \\ $$$$\:\:\:\:\:\frac{{a}}{\mathrm{sin}\:\left(\alpha+\phi\right)}=\frac{{b}}{\mathrm{sin}\:\phi}\:\:\:\:\:{and} \\ $$$${From}\:\bigtriangleup{BDE} \\ $$$$\:\:\:\:\:\:\frac{{c}}{\mathrm{sin}\:\alpha}\:=\:\frac{{b}}{\mathrm{sin}\:\left(\alpha+\beta\right)} \\ $$$$\mathrm{tan}\:\boldsymbol{\theta}\:=\:\frac{\mathrm{sin}\:\beta}{\frac{{a}}{{c}}\mathrm{cos}\:\phi−\frac{{b}}{{c}}\mathrm{cos}\:\mathrm{2}\beta+\mathrm{cos}\:\beta} \\ $$$$\:\:=\frac{\mathrm{sin}\:\beta}{\frac{\boldsymbol{{b}}\:\mathrm{sin}\:\left(\alpha+\phi\right)\mathrm{cos}\:\phi}{\boldsymbol{{c}}\:\mathrm{sin}\:\phi}−\frac{\boldsymbol{{b}}}{\boldsymbol{{c}}}\mathrm{cos}\:\mathrm{2}\boldsymbol{\beta}+\mathrm{cos}\:\boldsymbol{\beta}} \\ $$$$\:\:=\frac{\mathrm{sin}\:\boldsymbol{\beta}}{\frac{\mathrm{sin}\:\left(\alpha+\beta\right)}{\mathrm{sin}\:\alpha}\left[\frac{\mathrm{sin}\:\left(\alpha+\phi\right)\mathrm{cos}\:\phi}{\mathrm{sin}\:\phi}−\mathrm{cos}\:\mathrm{2}\beta\right]+\mathrm{cos}\:\beta} \\ $$$$\alpha+\phi\:=\:\mathrm{42}°+\mathrm{30}°\:=\mathrm{72}°\:=\:\mathrm{2}\beta\:\:,\:{so} \\ $$$$=\frac{\mathrm{sin}\:\boldsymbol{\beta}}{\frac{\mathrm{sin}\:\left(\alpha+\beta\right)}{\mathrm{sin}\:\alpha}\left[\frac{\mathrm{sin}\:\left(\mathrm{2}\beta−\phi\right)}{\mathrm{sin}\:\phi}\right]+\mathrm{cos}\:\beta} \\ $$$$\:\:\:\:\mathrm{2}\beta−\phi\:=\:\alpha \\ $$$$=\frac{\mathrm{sin}\:\beta}{\frac{\mathrm{sin}\:\left(\alpha+\beta\right)}{\mathrm{sin}\:\phi}+\mathrm{cos}\:\beta} \\ $$$$\:\:\Rightarrow\:\mathrm{tan}\:\theta\:=\:\frac{\mathrm{sin}\:\mathrm{36}°}{\mathrm{2sin}\:\mathrm{78}°+\mathrm{cos}\:\mathrm{36}°} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{sin}\:\mathrm{12}°\left(\mathrm{3}−\mathrm{4sin}\:^{\mathrm{2}} \mathrm{12}°\right)}{\mathrm{2cos}\:\mathrm{12}°+\mathrm{cos}\:\mathrm{12}°\left(\mathrm{4cos}\:^{\mathrm{2}} \mathrm{12}°−\mathrm{3}\right)} \\ $$$$=\:\mathrm{tan}\:\mathrm{12}°\:\left[\frac{\mathrm{3}−\mathrm{4sin}\:^{\mathrm{2}} \mathrm{12}°}{\mathrm{2}+\mathrm{4}−\mathrm{4sin}\:^{\mathrm{2}} \mathrm{12}°−\mathrm{3}}\right] \\ $$$$\mathrm{tan}\:\boldsymbol{\theta}\:=\:\mathrm{tan}\:\mathrm{12}°\: \\ $$$$\Rightarrow\:\:\:\boldsymbol{{x}}\:=\:\boldsymbol{\theta}\:=\:\mathrm{12}°\:\:\:. \\ $$
Commented by naka3546 last updated on 18/Feb/18
$${yes}\:\:\mathrm{12}° \\ $$