# two different values for angle A ?

• May 20th 2010, 05:10 PM
misspalmer
two different values for angle A ?
Okay so here is the question

Each angle of A is between 0 degrees and 180 degrees. Which equations result in two different values for angle A ?

-sin A = .7071
-cos A = -.5
-sin A = .9269
-cos A = -.7071
-sin A = .8660
-cos A = -1
-sin A = 3/4
-cos A = 3/4
-cos A = -3/4

I would like how you got there as well ! thanks !
• May 20th 2010, 10:55 PM
Hello misspalmer

Welcome to Math Help Forum!
Quote:

Originally Posted by misspalmer
Okay so here is the question

Each angle of A is between 0 degrees and 180 degrees. Which equations result in two different values for angle A ?

-sin A = .7071
-cos A = -.5
-sin A = .9269
-cos A = -.7071
-sin A = .8660
-cos A = -1
-sin A = 3/4
-cos A = 3/4
-cos A = -3/4

I would like how you got there as well ! thanks !

First, I assume that the hyphens at the left-hand end of each equation are not minus signs. You do need to be careful in writing down the questions!

Then you need to know how sine and cosine behave between $\displaystyle 0^o$ and $\displaystyle 180^o$. So take a look at the diagram I've attached.

You'll see that $\displaystyle \sin x$ goes from $\displaystyle 0$ to $\displaystyle 1$ as $\displaystyle x$ goes from $\displaystyle 0^o$ to $\displaystyle 90^o$. Then, in a symmetrical way, it goes back down to $\displaystyle 0$ as $\displaystyle x$ goes from $\displaystyle 90^o$ to $\displaystyle 180^o$.

So, for any number between $\displaystyle 0$ and $\displaystyle 1$, there will be two angles whose sine is this number: one between $\displaystyle 0^o$ and $\displaystyle 90^o$ and one between $\displaystyle 90^o$ and $\displaystyle 180^o$. But if the number lies outside the range $\displaystyle 0$ to $\displaystyle 1$, there won't be any angles between $\displaystyle 0^o$ and $\displaystyle 180^o$ having this number as their sine.

On the other hand, $\displaystyle \cos x$ starts at $\displaystyle 1$ and goes down to $\displaystyle 0$ as $\displaystyle x$ goes from $\displaystyle 0^o$ to $\displaystyle 90^o$. Then it continues down to $\displaystyle {-1}$ as $\displaystyle x$ goes from $\displaystyle 90^o$ to $\displaystyle 180^o$.

So no value of $\displaystyle \cos x$ is repeated as $\displaystyle x$ takes values between $\displaystyle 0^o$ and $\displaystyle 180^o$. So none of the equations involving cosine will have two answers.

If you apply the rules I've just given you, it should be pretty clear what the answers are. For instance, the first equation, $\displaystyle \sin A=0.7071$, will have two solutions, but the second, $\displaystyle \cos A = -0.5$, won't.

Can you do the rest of them now?