1. ## Coterminal Angles

Hello

someone help me with these problems please, ASAP

Which angle is coterminal with 38 degree 50' ?
a)-398 degrees 50'
b)-321 degrees 50'
c) 321 degrees 10'
d) 398 degrees 10'
c) none of these

---------------------------------

Find the exact values of 0 degrees and 360 degrees that make the following equation a true statement:
tan theta= -square root of 3 divided by3

2. Hello,

A and B are coterminal angles if A-B=360.

Thus it has to be 321 or -321. According to the definition, it'll be -321 degrees 50', since 38-(-321) ~ 360

The problem is that when we are talking about degrees ', the ' corresponds to minutes of degrees. 1 degree makes 60' (1 degree is like an hour). So it can't be -321°50', since 50'+50' $\neq$ 1°=60'.

So it's none of the solutions.

~~~~~~~~~~~~

$\tan(\theta)=-\frac{\sqrt{3}}{3}=-\frac{\sqrt{3}}{\sqrt{3} \cdot \sqrt{3}}=-\frac{1}{\sqrt{3}}$

$\tan(\theta)={\color{red}-\frac 12} \cdot \frac{2}{\sqrt{3}}=\frac 12 \cdot {\color{red}-\frac{2}{\sqrt{3}}}$

Can you continue ?

3. Originally Posted by Moo
Hello,

A and B are coterminal angles if A-B=360.

Thus it has to be 321 or -321. According to the definition, it'll be -321 degrees 50', since 38-(-321) ~ 360

The problem is that when we are talking about degrees ', the ' corresponds to minutes of degrees. 1 degree makes 60' (1 degree is like an hour). So it can't be -321°50', since 50'+50' $\neq$ 1°=60'.

So it's none of the solutions.

~~~~~~~~~~~~

$\tan(\theta)=-\frac{\sqrt{3}}{3}=-\frac{\sqrt{3}}{\sqrt{3} \cdot \sqrt{3}}=-\frac{1}{\sqrt{3}}$

$\tan(\theta)={\color{red}-\frac 12} \cdot \frac{2}{\sqrt{3}}=\frac 12 \cdot {\color{red}-\frac{2}{\sqrt{3}}}$

Can you continue ?
umm no i dont get the second problem

4. Originally Posted by subzero06
umm no i dont get the second problem
Look at the result that Moo gave you:

$\tan(\theta)=\frac12\left(-\frac{2}{\sqrt{3}}\right)$

This is really the same as saying

$\displaystyle{\tan(\theta)=\frac{-\frac12}{\frac{\sqrt{3}}2}}$

You should know, from the basic trigonometric identities, that $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$, so what value for $\theta$ gives a sine of $-\frac12$ and a cosine of $\frac{\sqrt{3}}2$?

5. Originally Posted by Reckoner
Look at the result that Moo gave you:

$\tan(\theta)=\frac12\left(-\frac{2}{\sqrt{3}}\right)$

This is really the same as saying

$\displaystyle{\tan(\theta)=\frac{-\frac12}{\frac{\sqrt{3}}2}}$

You should know, from the basic trigonometric identities, that $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$, so what value for $\theta$ gives a sine of $-\frac12$ and a cosine of $\frac{\sqrt{3}}2$?
Ok,
so the values for cosine is 30 degrees and sine -30 because is -(1/2)???

6. Originally Posted by subzero06
Ok,
so the values for cosine is 30 degrees and sine -30 because is -(1/2)???
Hm...
Let's take the first situation, OK ?

$\displaystyle{\tan(\theta)=\frac{-\frac12}{\frac{\sqrt{3}}2}}$

So it means that $\sin(\theta)=-\frac 12$ and $\cos(\theta)=\frac{\sqrt{3}}{2}$

What is the angle for which these two conditions are available ?

The other one is :

$\displaystyle{\tan(\theta)=\frac{\frac12}{-\frac{\sqrt{3}}2}}$

So it means that $\sin(\theta)=\frac 12$ and $\cos(\theta)=-\frac{\sqrt{3}}{2}$

7. Originally Posted by Moo
The other one is :
There are indeed two possible values on the given interval. That must have slipped my mind; thanks.

8. Originally Posted by Reckoner
There are indeed two possible values on the given interval. That must have slipped my mind; thanks.
Actually, that's why I put two equalities with two red things ^^

9. Lol, if you want the answers, I've just found it in google :