• Matéria: Matemática
  • Autor: sminnie753
  • Perguntado 4 anos atrás

me ajudem a achar o valor de x ( trigonometria) ​

Anexos:

Respostas

respondido por: elizeugatao
1

\displaystyle \underline{\text{Aplicando seno}}: \\\\ \text{sen}(60^\circ)=\frac{25}{\text x} \\\\\\ \frac{\sqrt{3}}{2}=\frac{25}{\text x} \\\\\\ \text x = \frac{2.25}{\sqrt{3}} \to \text x =\frac{50}{\sqrt{3}}.\frac{\sqrt{3}}{\sqrt{3}}\to \text x = \frac{50\sqrt{3}}{(\sqrt3)^2}\\\\\\ \huge\boxed{\text x= \frac{50\sqrt{3}}{3}\ }\checkmark


sminnie753: obrigada
Perguntas similares