Triângulos - 2008 - Vol. 2 ⇒ 020 - Triângulos - 2008 Tópico resolvido

[petrasMOD]

Do gráfico, calcular x, se: AB = BC = BD

Resposta

---

Gabarito: 7,5^o

[petrasMOD]

Seja E o vértice do ângulo x
[tex3]\mathsf{

\angle CAB = \angle CBA = \frac{180-30}{2} = 75^o \implies 2\beta +\angle CAD = 75^o \implies \angle CAD = 75^o - 2\beta (I)\\
\angle BCD = \angle CDB = 180 -75 - 2\theta = 105^o -2\theta(II)\\
\angle CBD = 180-2(105-2\theta) = 4\theta - 30^o (III)\\
\triangle ABE: x+ \beta+75+105 - 2\theta + \theta = 180 \implies x = \theta - \beta(IV) \\
\triangle ACD:\angle CDA = 180-\angle DAB- 30- \angle CBD = 180-(75-2\beta) -30 - (4\theta -30)\\
\therefore \angle CDA = 105-4\theta +2\beta\\
\angle CAD = \angle CDA \implies75-2\beta = 105-4\theta+2\beta \implies \theta - \beta = \frac{15}{2}\\
De(IV): \boxed{x=\frac{15}{2} = 7,5^o }

}[/tex3]