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Question Number 30160 by naka3546 last updated on 17/Feb/18
Commented by ajfour last updated on 18/Feb/18
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)
Commented by naka3546 last updated on 18/Feb/18
