# Thread: Basic Problem, need help.

1. ## Basic Problem, need help.

In my analysis of functions class we just started basic trig and have gone over the 6 trig functions and I understand them. Just a little confused on this problem my teacher assigned us over the weekend.

the directions include to make an illistration of the problem and how we solved it.

"A telephone pole is 60 feet tall. A guy wire 75 feet long is attached from the ground to the top of the pole. Find the angle between the wire and the pole to the nearest degree."

using the pythagorean theorm I found the adj side is 45 feet. I need help finding the angle:

2. Originally Posted by zbrownleep7
In my analysis of functions class we just started basic trig and have gone over the 6 trig functions and I understand them. Just a little confused on this problem my teacher assigned us over the weekend.

the directions include to make an illistration of the problem and how we solved it.

"A telephone pole is 60 feet tall. A guy wire 75 feet long is attached from the ground to the top of the pole. Find the angle between the wire and the pole to the nearest degree."

using the pythagorean theorm I found the adj side is 45 feet. I need help finding the angle:

SOH-CAH-TOA;

You don't even need to use the Pythag. Thm. Use cosine;

Does that help?

3. How does that give me the angle between the pole and wire?

Do i do 60/75?

4. Originally Posted by zbrownleep7
How does that give me the angle between the pole and wire?

Do i do 60/75?
cos(theta)=60/75=4/5=0.8,

now look up what angle has a cosine of 0.8 in a table, or use the
arccos (or acos or cos^-1 depending on your calculator) function on your

RonL

5. Hello, zbrownleep7!

A telephone pole is 60 feet tall.
A guy wire 75 feet long is attached from the ground to the top of the pole.
Find the angle between the wire and the pole to the nearest degree.
Code:
                              *
*  |
*  θ  |
*        |
75 *           | 60
*              |
*                 |
*                    |
* - - - - - - - - - - - +

We want angle $\theta$ in the right triangle.
Its adjacent side is 60 and the hypotenuse is 75.

Since $\cos\theta = \frac{adj}{hyp}$, we have: . $\cos\theta \:=\:\frac{60}{75} \:=\:0.8$

The angle whose cosine is 0.8 is: . $\theta \:=\:\cos^{-1}(0.8) \:=\:36.86989765^o$

Therefore: . $\theta \:\approx\:37^o$

6. The angle whose cosine is 0.8 is: . $\theta \:=\:\cos^{-1}(0.8) \:=\:36.86989765^o$

Therefore: . $\theta \:\approx\:37^o$

[/size]
Thank You! All I need to know is what the process is called when you raised the cos to -1 to get 36.8, is that the sec, inverse of cos? Appreciate it!

7. Originally Posted by zbrownleep7
Thank You! All I need to know is what the process is called when you raised the cos to -1 to get 36.8, is that the sec, inverse of cos? Appreciate it!
The notation IS confusing isn't it?

Here's a quick way to tell. When you take the cosine (or any trig function) of an angle it gives you a number. When you take the inverse cosine of a number it gives you an angle. Since Soroban's answer is in degrees, it must be the inverse cosine.

-Dan

8. Originally Posted by topsquark
Since Soroban's answer is in degrees, it must be the inverse cosine.

-Dan
So, if I'm not mistaken its just another way of getting the secant, correct?

9. Originally Posted by zbrownleep7
So, if I'm not mistaken its just another way of getting the secant, correct?
No. $sec \theta = \frac{1}{cos \theta}$ gives a number as an answer. $cos^{-1}x$ is an inverse function and gives an angle, usually measured in radians, but often measured in degrees. They are very distinct functions with a similar notation, hence the confusion. Always look at what kind of quantity they give for an answer and you will be able to tell them apart.

Another notation which is older (and not in as much use, unfortunately) is the term "arccosine." $acs(x) = cos^{-1}(x)$ without all the fuss about what the "-1" is supposed to mean. I prefer this notation myself, but I learned inverse trig functions from an old Calculus book, which used it exclusively.

-Dan

10. Excellent! Thanks for the help!