Question

(Unifal-MG) Observe a figura, em que m(B) = 60°, m(C) = 45° e AB = 2 m.
Determine a medida a no triângulo ABC.

Answer 1

Resposta:

2,732

Explicação passo-a-passo:

a = BD + DC

sen60 = h/2

√3/2 = h/2

h = √3

cos60 = BD/2

1/2 = BD/2

BD = 1

tg45 = h/DC

1 = √3/DC

DC = √3

a = 1 + √3

a = 1 + 1,732

a = 2,732

Answer 2

No ∆ABD:

$\mathsf{cos(60^{\circ})=\dfrac{\overline{BD}}{\overline{AB}}}\\\mathsf{\dfrac{1}{2}=\dfrac{\overline{BD}}{2}}$

$\mathsf{\overline{BD}=2.\dfrac{1}{2}=1m}$

$\mathsf{tg(60^{\circ})=\dfrac{\overline{AD}}{\overline{BD}}}\\\mathsf{\sqrt{3}=\dfrac{AD}{1}}\\\mathsf{\overline{AD}=\sqrt{3}}$

O ∆ADC é isósceles portanto $\mathsf{\overline{AD} =\overline{DC}=\sqrt{3}}$

$\mathsf{a=\overline{BD}+\overline{DC}}$

$\boxed{\boxed{\boxed{\mathsf{a=(1+\sqrt{3})m}}}}$