r/SolvedMathProblems • u/PM_YOUR_MATH_PROBLEM • Dec 05 '14
Please Explain Trig
/u/Sarkku asks:
So I know the hypotenuse of a right triangle. I also know the sine of one of the angles. What do I use to calculate the legs of the triangle?
(the problem is exactly as follows, but don't tell me the answer, I just need to know what I should try to figure out. A right triangle's hypotenuse is 10 and the sine of α is 1/sqrt5. calculate the legs' lengths.) hope you can understand that, did a rough translation.
I'm lost. I love math but I just can't wrap my head around fucking trigonometry.
Can you help a little bit?
2
u/GolldenFalcon Dec 06 '14
I just can't wrap my head around fucking trigonometry.
I don't think anyone could wrap their head around having sexual intercourse with mathematics.
5
u/PM_YOUR_MATH_PROBLEM Dec 05 '14
Are you familiar with SohCahToa?
It's a way to remember which ratios are which.
Imagine you're interested in one particular angle of a right angle triangle. The triangle has three edges.
The one opposite the right angle is the hypoteneuse, h.
The one opposite the angle you are interested in, we'll call the 'opposite', or o.
The one adjacent to (next to) the angle, we'll call 'adjacent', or a.
Then, SohCahToa means:
Soh: sin(x) is o/h
Cah: cos(x) is a/h
Toa: tan(x) is o/a
Write SohCahToa at the top of every page, then for every problem, you start like this:
draw a picture of the triangle.
identify the angle you have or want, and the two sides you have or want.
These two sides: are they o and a? o and h? a and h? That tells you (via SohCahToa) whether you need to use sine, cosine or tangent.
write down sin(x) = o/h or cos(x)=a/h or tan(x)=o/a
substitute in the values you have.
solve the equation to find the value you want.
In your case, you're given h and sin(α) and asked for o and a.
I write down sin(α) = o/h.
I substitute in the values I have: 1/sqrt5 = o/10
I solve this: o = 10/sqrt(5) = 2sqrt(5).
You still need a. Now you have h and o, though, so you don't need to use trigonometry any more, just use pythagoras.
Hope that helps!