Question

Num triângulo ABC, o ângulo A= 60o, o ângulo B= 45o e BC = 4. Calcule AC.

Answer 1

Lei dos senos:

$\frac{a}{ \sin(a) } = \frac{b}{ \sin(b) } \\ \\ \frac{4}{ \sin(60 {}^{o} ) } = \frac{AC}{ \sin(45 {}^{o} ) } \\ \\ \frac{4}{ \frac{\sqrt{3}}{2} } = \frac{AC}{ \frac{ \sqrt{2} }{2} } \\ \\ 4 \times \frac{2}{\sqrt{3}} = AC \times \frac{2}{ \sqrt{2} } \\ \\ \frac{8}{\sqrt{3}}= \frac{2AC}{ \sqrt{2} } \\ \\ 2×\sqrt{3}AC = 8 \sqrt{2} \\ \\ AC = \frac{8 \sqrt{2} }{2\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}} \\ \\ \frac{4\sqrt{2×3}}{\sqrt{3×3}} \\\\\frac{4\sqrt{6}}{\sqrt{9}}\\ \\ AC = \frac{4 \sqrt{6} }{3}$

Bons Estudos!