Question

If the sin 60 agrees equals the square root of three over two which statement is true

Answer 1

Given that
$\sin60^o= \frac{ \sqrt{3} }{2}$

$\cos 30^o =\sin(90-30)^o=\sin60^o=\frac{ \sqrt{3} }{2}$,

$\cos 120^o =-\sin(180-120)^o=-\sin60^o=-\frac{ \sqrt{3} }{2}$,

$\cos30^o\neq0$,

$\cos120^o\neq0$

Therefore, the true statement is
$\cos 30^o =\sin(90-30)^o=\sin60^o=\frac{ \sqrt{3} }{2}$,
because the cosine and sine are complements.