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Precalculus 15 Online
OpenStudy (anonymous):

if cos(t)=-1/5 and pi

zepdrix (zepdrix):

|dw:1362772905106:dw|Understand why I drew that triangle in the third quadrant?

OpenStudy (anonymous):

not quite sure why?

zepdrix (zepdrix):

|dw:1362773055192:dw| They told us that the angle t in within this interval.\(\large \pi<t<3\pi/2\). So we draw a line in the third quadrant and form a triangle from it.

OpenStudy (anonymous):

ok now i understand that part

zepdrix (zepdrix):

|dw:1362773145892:dw|Let's look at just the triangle now. Don't worry about the fact that I labeled the angle \(\large t'\). That's not super important. Just think of it as \(\large t\).

zepdrix (zepdrix):

\(\large \cos t=-\dfrac{1}{5}=\dfrac{adjacent}{hypotenuse}\) So how would we label these sides? HINT: We always let the hypotenuse be positive.

OpenStudy (anonymous):

|dw:1362773386830:dw|

zepdrix (zepdrix):

Ok looks good! Just need to find the missing side using the `Pythagorean Theorem` and then we just have one small step after that.

OpenStudy (anonymous):

opposite will equal 4?

OpenStudy (anonymous):

sorry did an error......missing side equals to \[+/- 2\sqrt{6}\]

zepdrix (zepdrix):

Ok looks good. Now remember that we drew our triangle in the `third` quadrant. So for that OPPOSITE leg, did we move UP or DOWN? Do we want the positive or negative answer?

OpenStudy (anonymous):

we want a negative answer because tan is only positive in the 3rd Q

zepdrix (zepdrix):

Haha that's an interesting way to think about it XD I was thinking about the fact that we moved downward in the y direction to make the opposite length. So it would be negative. But whatever works for you! :D|dw:1362773803586:dw|

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