Solving for a Hypotenuse: A Frustrating Problem for Morgan

[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

 

[happyg1]

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

 

[rockmorg]

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

 

[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

 

[happyg1]

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

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.

 

[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!