Question

If cos Θ = negative two over five and tan Θ > 0, what is the value of sin Θ?

negative square root 21

negative square root 21 over 5

square root 21 over 5

square root of 21 over 2

Answer 1

\[a^2 + b^2 = c^2\]\[(-2)^2 + x^2 = 5^2\]\[x^2 = 25 - 4\]\[x = \sqrt{21}\]
\[\sin(Θ) = x/5\]\[\sin(Θ) = -\sqrt{21}/5\]

Answer 2

So, negative square root 21 over 5

Answer 3

so you just plugged the into the pythagoren theorem, I get it now :) ty

Answer 4

\(\bf cos(\theta)=-\cfrac{2}{5}\qquad tan(\theta)>0
\\ \quad \\
tan(\theta)>0\Leftarrow\textit{another way of saying, tangent is positive}\\
\textit{meaning, 3rd or 1st Quadrants}
\\ \quad \\
cos(\theta)=-\cfrac{2}{5}\implies \cfrac{adjacent}{hypotenuse}\implies \cfrac{a=-2}{c=5}\)

notice ASNinja 's picture coordinates