Question

Me ajudem nessa questão de triângulo
retângulo

Answer 1

Resposta:

$\sin(45) = \frac{x}{22 \sqrt{2} } \\ 22 \sqrt{2} = \frac{x}{ \sin(45) } \\ 22 \sqrt{2} = \frac{x}{ \frac{ \sqrt{2} }{2} } \\ \frac{ \sqrt{2} }{2} (22 \sqrt{2}) = x \\ 11 \sqrt{2} \sqrt{2} = x \\ 11 \times 2 = x \\ x = 22$

$\cos(45) = \frac{y}{22 \sqrt{2} } \\ \frac{ \sqrt{2} }{2} = \frac{y}{22 \sqrt{2} } \\ 22 \sqrt{2} ( \frac{ \sqrt{2} }{2} ) = y \\ y = 11 \sqrt{2} \sqrt{2} \\ y = 11 \times 2 \\ y = 22$