Geometria Plana

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Obs.: ( Â + Â' = pi ) Acredito que deve ter sido digitado errado.

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i) Pela lei dos senos temos:

[latex]\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R[/latex]

[latex]\frac{a'}{\sin A'}=\frac{b'}{\sin B'}=\frac{c'}{\sin C'}=2R'[/latex]

Portanto temos:

[latex]c\cdot c'=(2\cdot R\cdot \sin {C})\cdot (2\cdot R'\cdot \sin {C'})\Rightarrow \frac{c\cdot c'}{4\cdot R\cdot R'}=\sin {C}\cdot \sin {C'}[/latex]



ii)Agora observe que:

[latex]\sin {C}=\sin {(\pi -A-B)}=\sin {(A+B)}[/latex]

e

[latex]\sin {C'}=\sin {(\pi -A'-B')}=\sin {(\pi -(\pi -A)-B)}=\sin {(A-B)}[/latex]

Logo,

[latex]\frac{c\cdot c'}{4\cdot R\cdot R'}=\sin {C}\cdot \sin {C'}=\sin {(A+B)}\cdot \sin {(A-B)}=\frac{\cos 2B - \cos 2A}{2}[/latex]

[latex]\frac{c\cdot c'}{4\cdot R\cdot R'}=\frac{\cos 2B - \cos 2A}{2}=\frac{(1-2\cdot \sin ^2B)-(1-2\cdot \sin ^2A)}{2}[/latex]

[latex]\frac{c\cdot c'}{4\cdot R\cdot R'}=\sin ^2A - \sin ^2B\Rightarrow c\cdot c'=4\cdot R\cdot R'\cdot \sin ^2A - 4\cdot R\cdot R'\cdot \sin ^2B[/latex]

[latex]c\cdot c'=a\cdot a' - b\cdot b'[/latex]

[latex]a\cdot a' = b\cdot b' + c\cdot c'[/latex]