Coterminal Angles

**subzero06** · May 3rd 2008, 07:57 PM

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

**Moo** · May 3rd 2008, 11:36 PM

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 ?

**subzero06** · May 4th 2008, 11:03 AM

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

**Reckoner** · May 4th 2008, 11:52 AM

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 $\frac{\sqrt{3}}2$ ?

**subzero06** · May 4th 2008, 12:15 PM

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)???

**Moo** · May 4th 2008, 12:23 PM

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 $\cos(\theta)=-\frac{\sqrt{3}}{2}$

**Reckoner** · May 4th 2008, 12:34 PM

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.

**Moo** · May 4th 2008, 12:36 PM

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 ^^

**Moo** · May 4th 2008, 12:54 PM

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

Thread: Coterminal Angles

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