Cosine Problem

1. Oct 3, 2005

rockmorg

Hey all -

I'm going insane because I can't think of how to do this...

I can't for the life of me get this straight... I have a triangle that I'm trying to solve for the hypotenuse, angle theta is 70 degrees and I know the adjacent leg is 7500. Cosine SHOULD solve for the hypotenuse, right (adjacent/hypotenuse)? I can't get it to solve for the hypotenuse.

I set it up like this:

cos70 = 7500/y

I need to get the 7500 on the side with the cos70, but wouldn't I have to multiply the right by 1/7500 to cancel it on that side for y (hypotenuse)?

Any help would be appreciated, I'm running into this over and over with my probelms.

Thanks!
-
Morgan

2. Oct 3, 2005

happyg1

Just invert both sides:
1/cos 70 = y/7500
then solve for y

3. Oct 3, 2005

rockmorg

I have to be missing something... because there is a similar problem where they basically (diff. numbers) make it (7500)cos70 = y

4. Oct 3, 2005

Ryan231

I'm no expert but I did it like this..
cos70 = 7500/x x= hypot

cos70 = .3420

.3420 = 7500 /x
.3420x = 7500
x = 7500 / .3420
x = 21929.82456

5. Oct 3, 2005

happyg1

OK
Here's another approach:
You have it set up correctly.
Multiply both sides by y to get the y out of the denominator on the RHS:
y*cos 70 = 7500
then solve for y.
The "similar problem" is not the same as this one. This is just a couple of fractions that are making your head hurt. Think about how you would solve this:
2 = 1/x
you can either take the inverse of both sides to get
1/2 = x
or you can multiply both sides by x to get
2x = 1
and then solve for x.
It's the exact same result.
hope this helps.

6. Oct 3, 2005

rockmorg

hmmm okay I'm seeing it is not the same as the other problem that I thought was the same... so your answers make sense.

thanks guys!