Question

Determine o valor de X

Answer 1

$a)cos(60) = \frac{cateto \: adjacente}{hipotenusa} \\ \frac{1}{2} = \frac{x}{26} \\ 2x = 26 \\ x = \frac{26}{2} \\ x = 13 \\ \\ b)tg(30) = \frac{cateto \: oposto}{cateto \: adjacente} \\ \frac{ \sqrt{3} }{3} = \frac{16}{x} \\ x \sqrt{3} = 3 \times 16 \\ x \sqrt{3} = 48 \\ x = \frac{48}{ \sqrt{3} } \\ x = \frac{ 48\sqrt{3} }{ \sqrt{3}. \sqrt{3} } = \frac{48 \sqrt{3} }{ \sqrt{9} } = \frac{48 \sqrt{3} }{3} \\ x = 16 \sqrt{3}$