Question

gente, qual o assunto de matematica dessa questao?​

Answer 1

No ∆AMC

$\tan(60) = \frac{h}{x} \\ \sqrt{3} = \frac{h}{x} \\ h = x \sqrt{3}$

No ∆ABC

$\tan(30) = \frac{h}{x + 20} \\ \frac{ \sqrt{3} }{3} = \frac{x \sqrt{3} }{x + 20} \\ 3x = x + 20 \\ 3x - x = 20$

$2x = 20 \\ x = \frac{20}{2} \\ x = 10$

$h = x \sqrt{3} \\ h = 10 \sqrt{3}$

Alternativa b