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Re: [obm-l] TRIANGULO ABC
- To: obm-l@xxxxxxxxxxxxxx
- Subject: Re: [obm-l] TRIANGULO ABC
- From: "Ivan lopes" <lopesivan.del@xxxxxxxxx>
- Date: Sat, 29 Sep 2007 05:33:02 -0300
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Questao:
(UFPB-77) Num triângulo ABC cujos ângulos são A, B e C. Supõe-se que
2tg A = tg B + tg C e 0 < A < pi/2. Neste triângulo vale a relação:
a) tg B.tg C = 3.
b) cos (B – C) = 2sec A.
c) cos (B + C) = 2cos A.
d) tg B.tg C = rq3.
e) nenhuma das respostas.
solucao:
2tgA = tgB + tgC (#eq1)
sendo Pi = A + B + C,
temos que
A = Pi - (B + C)
logo
tgA = tg[Pi - (B + C)]
-> tgA = - tg[B + C] (#eq2)
todavia, podemos escrever um
sistemacomposto por duas equacoes
2.tgA = tgB + tgC
tgA = - tg[B + C]
substituindo o valor de tgA obtido em (#eq2) em (#tgA), temos
2.tgA = tgB + tgC
-> -2.tg[B + C] = tgB + tgC
-> -2.(tgB + tgC)/(1-tgB.tgC) = tgB + tgC **dividindo por (tgB + tgC)
-> -2/(1-tgB.tgC) = 1
-> -2 = 1-tgB.tgC
-> -2 -1 = -tgB.tgC
-> tgB.tgC = 3
Resposta: (a)
by Ivan