[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
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
- DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=gmail.com; s=beta; h=domainkey-signature:received:received:message-id:date:from:to:subject:in-reply-to:mime-version:content-type:references; bh=oA7Iq9DPHpp07CKfddbfBWiXPm4W+72LZ/1NQdlbIFg=; b=rfJFJdLCHd066qiR0ODgwIPj3YWtOIGiDWSPJPYvc71jPiK1OFcugAMg++wRN/NYAvPg2ozcqhYKx6DaPNHugRAGdzKyut8oCqeIhY5ai9CBx+lV8UZPzRk2M67/PT6/L+PAWQR988QuOJTDTRk5ZcgrJ4gk9rUh9KXcZf4cyAQ=
- DomainKey-Signature: a=rsa-sha1; c=nofws; d=gmail.com; s=beta; h=received:message-id:date:from:to:subject:in-reply-to:mime-version:content-type:references; b=sxsZwQl/eCzz/DvzjymWDgohxL+VZ0unprNQC/powl/4lwal4lHuUBxwpst6mU2FQTobR4VtfoRdLHTgkdsZRpU3vhANHiYbOY5lty6A2yQ5XH59S3CHdcypt3mkKqTZetw7jcPFhZF/0v9VxuMp3mLTJe4/ONMGLZcLnQbP+1U=
- In-Reply-To: <JP2UI4$ACF1EF3A036EB9957C49A4389FE9C20F@bol.com.br>
- References: <JP2UI4$ACF1EF3A036EB9957C49A4389FE9C20F@bol.com.br>
- Reply-To: obm-l@xxxxxxxxxxxxxx
- Sender: owner-obm-l@xxxxxxxxxxxxxx
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