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/sci/ - Science & Math


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4295846 No.4295846 [Reply] [Original]

Hey there /sci/entists.
I need college math help.
How the FUCK do you find sides with the law of sine?
I cant figure this out at all.
>Here is an example.
<A=37 (degrees)
<C=95(degrees)
c=18m
Please explain this to me.

>> No.4295872

Definitely not college level.
I did this in my O levels.

>http://www.wolframalpha.com/input/?i=sin+%2837+deg%29+%2Fa+%3D+sin%2895+deg%29+%2F+18

>> No.4295877

>>4295872
Community college so thats probably why.

>> No.4295878

The ratio between the sine of an angle and the length of the opposite side is the same for all three angles in a triangle.

sin A / a = sin B / b = sin C / c

What don't you understand, specifically?

>> No.4295881

>>4295878
If i divide that I just get a number and a decimal, how does that correlate to finding the next side and such?

>> No.4295886

Fucking trig. This is middle school shit. Fine, I'll show you.

Law of sines is a ratio for any real triangle:
<div class="math">\frac{\sin(A)}{a}=\frac{\sin(B)}{b}=\frac{\sin(C)}{c}</div>
Plug in values:
<div class="math">\frac{\sin(37)}{a}=\frac{\sin(95)}{18}</div>
Solve for a:
<div class="math">\sin(37)=\frac{\sin(95)}{18}a</div>
<div class="math">\frac{18\sin(37)}{\sin(95)}=a</div>
Use your calculator, make sure its set to degrees.

>> No.4295893

>>4295886
Alot of help.
From there how do i get the rest of the sides?

>> No.4295906

>>4295893

A triangle has 180 degrees total. You can find B by doing <span class="math">180=A+B+C=37+B+95[/spoiler], or just subtracting (37+95) from 180. That's B, now use the law of sines to find the side lengths <span class="math">a,b[/spoiler].

Remember that the law of sines states that all three ratios are equal. You can set whichever one you like to the other so solve for the missing variable. Just make sure you know what variables you have and which ones you're aiming to solve for.

>> No.4295917

>>4295906
>>4295886
>>4295878
>>4295872
Thanks for everyones help. I can go on without being a retarded niggerfaggot.