posted by Jenny Lynn .
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?
Use SOH CAH TOA.
sinx = opposite/hypotenuse
cosx = adjacent/hypotenuse
tanx = opposite/adjacent
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?
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.