Question

Pergunta Trigonometria

Answer 1

Relações trigonométricas:

$\bullet\;\;\mathrm{sen\,}\alpha=\dfrac{\text{cateto oposto a }\alpha}{\text{hipotenusa}}\\ \\ \\ \bullet\;\;\cos \alpha=\dfrac{\text{cateto adjacente a }\alpha}{\text{hipotenusa}}\\ \\ \\ \bullet\;\;\mathrm{tg\,}\alpha=\dfrac{\text{cateto oposto a }\alpha}{\text{cateto adjacente a }\alpha}=\dfrac{\mathrm{sen\,}\alpha}{\cos \alpha}$


a) $\mathrm{tg\,}30^{\circ}=\dfrac{x}{20}$

$\dfrac{\sqrt{3}}{3}=\dfrac{x}{20}\\ \\ 3x=20\sqrt{3}\\ \\ x=\dfrac{20\sqrt{3}}{3}\\ \\ \\ \cos 30^{\circ}=\dfrac{20}{y}\\ \\ \dfrac{\sqrt{3}}{2}=\dfrac{20}{y}\\ \\ \sqrt{3}y=2\cdot 20\\ \\ y=\dfrac{40}{\sqrt{3}}\\ \\ y=\dfrac{40\sqrt{3}}{\sqrt{3}\cdot \sqrt{3}}\\ \\ y=\dfrac{40\sqrt{3}}{3}$


b) $\mathrm{sen\,}60^{\circ}=\dfrac{x}{10}$

$\dfrac{\sqrt{3}}{2}=\dfrac{x}{10}\\ \\ 2x=10\sqrt{3}\\ \\ x=\dfrac{10\sqrt{3}}{2}\\ \\ x=5\sqrt{3}\\ \\ \\ \cos 60^{\circ}=\dfrac{y}{10}\\ \\ \dfrac{1}{2}=\dfrac{y}{10}\\ \\ 2y=1\cdot 10\\ \\ y=\dfrac{10}{2}\\ \\ y=5$