Vol. 09 - Geometria Plana FME 2013 ⇒ 113 - FME 09 -Teste de Vestibulares 2013 Tópico resolvido

[petrasMOD]

(UF-BA) Na figura abaixo, tem-se:
[tex3]\angle[/tex3]BAC = 45º
[tex3]\angle[/tex3]BDC = 60º
AD = 5 u.c.
DC = 10 u.c.
Com base nesses dados, calcule BC.

Resposta

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Gabarito: BC = 5[tex3]\sqrt{6}[/tex3] u.c.

[petrasMOD]

[tex3]\mathsf{
\frac{BD}{sen45^o} = \frac{AD}{sen15^o } \implies \frac{x}{5} = \frac{\frac{\sqrt2}{2}}{\frac{\sqrt6-\sqrt2}{4}} \implies x = \frac{10\sqrt2}{\sqrt6-\sqrt2}\\
\therefore x = \frac{10\sqrt12-20}{4} = \frac{20\sqrt3+20}{4} =5\sqrt3+5\\
T.cossenos: BC^2 = 10^2+(5\sqrt3+5)^2-2.10.(5\sqrt3+5)(cos60^o ) =
\\100+75+50\sqrt3+25 - (100\sqrt3+100)(\frac{1}{2})\\
BC^2=200+50\sqrt3-50\sqrt3-50 = 150\\
\therefore BC = \sqrt{50} = \boxed{5\sqrt6}



}[/tex3]