Finding theta of right triangle

auslmar · #1 $Old$ June 15th, 2008, 03:34 PM

Hello,

I have a right triangle with hypotenuse x, adjacent side 4, and opposite side sqrt(x^(2) - 16) with angle theta.

I'm having trouble remembering trigonometric approaches to finding theta.

Previous parts of this problem involve integration using trigonometric substitution and I'm building the triangle to convert back to the original variable, x. It is very well possible that I have erred in doing that part of the problem, but this is where I got to, but couldn't proceed. I have the solution in terms of theta being, theta/16 + cosine(theta)sine(theta) + C.

Might I express theta as an inverse trig function, or is there a better way to do it?

If anyone has any pointers or tips, I'd greatly appreciate it.

Thanks for your consideration,

Austin Martin

kalagota · #2 $Old$ June 16th, 2008, 12:40 AM

i dont know if this would help you.. but the mnemonics we use is SOH-CAH-TOA

for a given right triangle with $\theta$ as an interior (acute) angle, then

$\sin \theta = \frac{Opposite}{Hypotenuse}$
$\cos \theta = \frac{Adjacent}{Hypotenuse}$
$\tan \theta = \frac{Opposite}{Adjacent}$

therefore, you can find $\theta$ using the inverse trigo function..