Question

podemm me ajudar???
preciso das respostas e calculos

Answer 1

hipotenusa=a
cateto adjacente=9

cos30°=$\frac{9}{a}$

$\frac{ \sqrt{3} }{2} = \frac{9}{a}$

$a \sqrt{3} =18$

$a= \frac{18}{ \sqrt{3} }$

vamos racionalizar

$x= \frac{18 \sqrt{3} }{ \sqrt{3}. \sqrt{3} } = \frac{18 \sqrt{3} }{ \sqrt{3^2} } = \frac{18 \sqrt{3} }{3} = 6 \sqrt{3}$

$\fbox{ a=6 \sqrt{3} }$

hipotenusa=a=6√3
cateto oposto=c

sen30°=$\frac{c}{a}$

sen30°=$\frac{c}{6 \sqrt{3} }$

$\frac{1}{2} = \frac{c}{6 \sqrt{3} }$

$2c=6 \sqrt{3}$

$c= \frac{6 \sqrt{3} }{2}$

$\fbox{ c=3 \sqrt{3} }$