If sintheta=2/3 which of the following are possible? sectheta= -3/2 and tantheta= 2/root5 costheta= -root5/3 and tantheta= 2/root5 costheta= root5/3 and tantheta= 2/root5 sectheta= 3/root5 and tantheta= 2/root5

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If sintheta=2/3 which of the following are possible? sectheta= -3/2 and tantheta= 2/root5 costheta= -root5/3 and tantheta= 2/root5 costheta= root5/3 and tantheta= 2/root5 sectheta= 3/root5 and tantheta= 2/root5

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First find the missing side. We know that sin\(\theta\)=\(\frac{2}{3}\), so the opposite side is \(2\), and the hypotenuse is \(3\). Now we need to find the adjacent. Use the Pythagorean theorem. \(c^2=a^2+b^2\) \(3^2=2^2+b^2\) \(b^2=9-4\) \(b=\sqrt{5}\) So the adjacent side is \(\sqrt{5}\).
okay im with it so far

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So now we know the three sides: opposite: 2 adjacent: \(\sqrt {5}\) hypotenuse: 3 ---- We know that \(\huge cos\theta\)=\(\huge \frac {adjacent}{hypotenuse}\) so \(\huge cos\theta\)=\(\huge \frac {\sqrt{5} }{3}\)
Now that we have the sides which are 2,3, and -root5 we know that sin=-3/root5 cos=2/5 tan=3/2 sec=-root5/2 csc= -root5/3 and cot= 2/3
i didnt mean negative root 5 just root 5
but i also see that i messed up
Oops okay i had them in the wrong order then
And \(\tan\theta =\huge \frac {oposite}{adjacent}\), so \(\tan\theta =\huge \frac {2}{\sqrt{5}}\) or \(\huge \frac {2\sqrt{5}} {5}\) .
opposite*
So that means that the last two in my question are the best two options
Yes! :)

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