Posted by **Jenny Lynn** on Monday, November 26, 2007 at 6:42pm.

Hello! I need assistance in solving for a side for a right triangle in trigonometry. Can anyone give me the easiest way/formula how to go about solving it?

Thank you!

- Math -
**Michael**, Monday, November 26, 2007 at 6:48pm
Use SOH CAH TOA.

sinx = opposite/hypotenuse

cosx = adjacent/hypotenuse

tanx = opposite/adjacent

- Math -
**Jenny Lynn**, Monday, November 26, 2007 at 8:19pm
I've come across that before..

But the main objective I have difficulty with is when you use the proportion (ex.)

angle decimal (after you look it up on the sin, cos, or tan table)/1 and object you are looking for/adjacent

What do you do afterward? Or is that it?

- Math -
**Nico**, Sunday, January 27, 2008 at 11:21pm
If you are looking for the hypotenuse, then the equation that you should use is z=SQRT(x^2 + y^2), where both x and y are shorter than the z and ae adjacent to the right angly.

However, if you are looking for one of the legs(either one of the shorter sides), then you must use y = SQRT(z^2 - x^2), where the z is the hypotenuse and x is the other leg.

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