if cos0 <0 and sin0= 2/5, find the exact value of tan0

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if cos0 <0 and sin0= 2/5, find the exact value of tan0

Mathematics
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easiest way to do this is draw a triangle. think of sine as opposite over hypotenuse, so label the opposite side 2 and the hypotenuse 5. all you need is the other leg which you get by pythagoras. \[s^2+2^2=5^2\] \[s^2=5^2-2^2\] \[s=\sqrt{5^2-2^2}=\sqrt{25-4}=\sqrt{21}\]
yeah i got that
but how do I get the exact value?

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tan is opposite of adjacent, so you you get \[tan(\theta)=-\frac{5}{\sqrt{21}}\] negative because you are told cosine is negative.
that is the 'exact value' if you want a decimal approximation you have to use a calculator, but that is not "exact".
wouldn't i have to multiply by square root of 21 to get rid of the square root?
if you want to rationalize the denominator that is fine, but rather unnecessary unless you have a math teacher that insists that you do it. you just get \[-\frac{5 \sqrt{21}}{21}\] which doesn't look that much better. in fact it looks worse.
yes he does insist on it! I wish he didn't. I got everything im just checking all my answers if you don't mind?
take your time i will check back in a second.
heres a question im not sure about. Find the length of an arc cut off by a central angle of 1.5 radians in a circle of radius 6 inches.
radian measure is arc length divided by radius. so if we call the arc length a you have \[\frac{a}{6}=1.5\] \[a=6\times 1.5 = 9\]
and what is the formula we use for that one?
and, from a point on the ground, 238 ft from the foot of a vertical tower, the angle of elevation of the top of the tower is 43 degrees. what is the height of the tower to the nearest tenth?
u there?

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