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Re: [obm-l] trigonometria
- To: obm-l@xxxxxxxxxxxxxx
- Subject: Re: [obm-l] trigonometria
- From: "saulo nilson" <saulo.nilson@xxxxxxxxx>
- Date: Tue, 27 Mar 2007 22:10:24 -0300
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cosB = senA/2*senC
A nao pode ser 90 porque senao teriamos
sen45=1 imp0ossivel
B=90
senA/2*senC=0
cos(A/2-C)=cos(a/2+C)
A/2-C=A/2+C
C=0 nao pode
A/2-C=A/2+C+360 impossivel
C=90
cosB=senA/2
B=A/2=pi/4
ou
B+A/2=90
ficamos com
B=45=2C letraB
On 3/26/07, Julio Sousa <juliosousajr@gmail.com> wrote:
1) Sendo A, B, C ângulos internos de um triângulo retângulo, prove que:
senA + senB + senC = 4*cos(A/2)*cos(B/2)*cos(C/2)
2) Se num triângulo retângulo for satisfeita a igualdade cosB = senA/2*senC, existirá entre seus ângulos a relação
a) B = A+C
b) B = 2C
c) C = 2B
d) C = A - B
e) B = C
--
Atenciosamente
Júlio Sousa