1. ## Trig Problem

I am doing a section on trigonometric functions of any angle and I have no idea how to do the folowing problem can someone help me out?

2. You should be able to draw two "special" triangles and label their sides. These are the 30°-60°-90° triangle and the 45°-45°-90° triangle. (See Special right triangles - Wikipedia, the free encyclopedia for diagrams.) Note that
$\frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$ and
${\sqrt{2}} = \frac{\sqrt{2}}{1}$.

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3. So for part (a) the measure of the triangles sides would be 1, 2, and radical 3?
Like this:

And because COT is hypotenuse (radical 3) over opposite (1) the angle would be 30 °?

4. No, that's not right. Cotangent is adjacent of opposite. Furthermore, for the 30°-60°-90° triangle, the hypotenuse is $2$, not $\sqrt{3}$. Can you try again?

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5. So for adjacent of opposite being 1/ . That would make the 1 adjacent and the the opposite? Which would make it for 60°?

6. Yes. Now do the similar with part (b).

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