Question

If cos(theta) = 3/5, then sin(theta) can be what?

Answer 1

\[\sin \theta=\frac{\sqrt{5^{2}-3^{2}}}{5}\]

Answer 2

Using Pythagorean theorem to get the "opposite" side of the triangle.
And opposite divided by hypotenuse is sine.

Answer 3

4/5

Answer 4

its a 3-4-5 right angled triangle

Answer 5

\[\text{Sin}[\text{ArcCos}[3/5]]=\frac{4}{5} \]

Answer 6

\[\cos^2 x+\sin^2 x=1\]
so \[\sin^2 x=1-\cos^2 x\]

Answer 7

you ok with these answers?

Answer 8

Any valid method that comes up with 4/5 is fine and dandy.

Answer 9

When I realized opposite/hypothenuse, I was hit with a long "OHHHHHHHHHH!"

Answer 10

Rudy, how did you get your answer?

Answer 11

How do you know that it is sqrt{5^2 - 3^2}

Answer 12

That's the Pythagorean theorem.
\[a^{2}=b^{2}+c^{2}\]
Where a is the hypotenuse, b and c are the other two sides. Since you have the hypotenuse, just solve for b or c and you get:
\[b^{2}=a^{2}-c^{2}\]
and solving for b, you get:

\[b=\sqrt{a^{2}-c^{2}}\]

Answer 13

and 3,4 5 triangle 9these are lengths of the sides) is a right angled triangle because
5^2 = 3^2 + 4^2