# Thread: [Trigonometry] Trying to figure out where it goes wrong

1. ## [Trigonometry] Trying to figure out where it goes wrong

So I have a 2D-triangle with three different values.

I've figured out that by doing cos A = ((a^2)+(b^2)-(c^2))/2*a*b that the angle of A is 69 degrees.

But now I try to figure out the other angles by using sinA/a = sinB/b = sinC/c, which means that sinB = (sinA*b)/a

However, when I try to do it I get that sin69 = 0,93. Now I can't get the sinB = (sinA*b)/a to work since that means that when I insert the values I get (0,93*8)/5,2 = 1,27

Obviously I can't get an angle by sin^-1 1,27 since it's not a value between -1 and 1. So now I'm stuck and I need some help trying to figure out where I might've gone wrong. Anyone?

2. From what I can tell yuo've done everything right. try using the law of sines to find angle C then use B=180-A-C.

From what I can tell yuo've done everything right. try using the law of sines to find angle C then use B=180-A-C.
I tried, but b = 8,2 and c = 8, so it's still wrong.

4. Or I'm wrong.

You used the law of cosines wrong. To find A you would need to use this version: a^2=b^2+c^2-2*b*c*cosA. If you solve for A you get cosA=[b^2+c^2-a^2]/2bc

---------- Post added 2012-12-14 at 03:40 AM ----------

If you solve for A then you get A=37,41 degrees. It follows by using the law of sines B=73,38 degrees and C=69,20 degrees

5. First of all, that was angle C you got, not A. <C= 67 degrees.

Next, you're doing the law of sins wrong because of that. Do sin(69)*5.2/8 = sin(A)

It all equals out.
A= 37
B= 73
C= 69

And some decimals in there.

6. OK so first off you have 3 unknowns (the 3 angles A, B, and C), so we know that we need 3 independent equations. Lets use the law of cosines twice, and the sum of angles once.

The law of cosines for A can be defined as A=arcos[(b^2 + c^2 - a^2)/2bc] ---> A=arcos[(8.2^2 + 8^2 - 5.2^2)/(2*8.2*8)] = 37.42

The law of cosines for B can be defined similarly.. B=arcos[(a^2 + c^2 - b^2)/2ac] ---> B=arcos[(5.2^2 + 8^2 - 8.2^2)/(2*5.2*8)] = 73.38

Our last equation is the sum of angles. We know that the sum of the angles of a triangle is 180...

---> 180 = A + B + C ---> C = 180 - A - B, therefore C =180 - 37.42 - 73.38 ---> C=69.20

A = 37.42
B = 73.38 } all in degrees
C = 69.20

EDIT: We are gonna go on the great assumption that i remembered how to math properly.

7. Thanks guys, however, for my assignment I need to use the law of sinus as well as cosinus, so I try to figure out the angles by using the law of sinus.

But I think I got it figured out now, so thanks, saved my ass there. :P

8. next time if you don't know where you're wrong just take a look at the sketch...do you think alpha could possibly be >45° as according to your results? :P

i mean i know it's just a sketch...but this is a huge discrepancy.

9. Originally Posted by arynzerr
next time if you don't know where you're wrong just take a look at the sketch...do you think alpha could possibly be >45° as according to your results? :P

i mean i know it's just a sketch...but this is a huge discrepancy.
I did wonder how that made sense, but I absolutely missed the formula and instead thought that I had calculated wrong.

10. Originally Posted by Tomatketchup
Thanks guys, however, for my assignment I need to use the law of sinus as well as cosinus, so I try to figure out the angles by using the law of sinus.

But I think I got it figured out now, so thanks, saved my ass there. :P

I don't see why you would law of sines. It is incorrect to use law of sines because you run into the scenario where sinA=x. Therefore solving for A would be arcsin(x) or 180-arcsin(x). The sketch would make it obvious what the correct choice is, however it's wrong to use this method. If you are required to use only law of sines and cosines, use law of cosines a third time rather than sum of angles. C = arcos[(a^2 + b^2 - c^2)/2ab]

11. Originally Posted by spoofy cow
I don't see why you would law of sines. It is incorrect to use law of sines because you run into the scenario where sinA=x. Therefore solving for A would be arcsin(x) or 180-arcsin(x).
This might seem like a stupid question, but why is it bad if sinA=x?

12. No idea what this guy is talking about to be honest. Go ahead and use the law of sines.

Also I don't know why he's saying not to use the sum of the angles when he used it himself in his earlier post lol...

Anyway you have the answer, just be mindful that you're actually solving for the angle you're doing the work for.

13. cos(a)=((8.2^2)+(8^2)-(5.2^2))/2(8.2)(8)
You find cos(a)=0.794207, then A=37.42°
sin(a)=sqrt(1-0.794207^2)=0.607647
Then from
(a/sin(a))=(b/sin(b)) you get
sin(b)=(b(sin(a)))/a
sin(b)=0.958212577, B=73.38°
C=180°-73.38°-37.42°=69.2°

14. Originally Posted by Blueobelisk
No idea what this guy is talking about to be honest. Go ahead and use the law of sines.

Also I don't know why he's saying not to use the sum of the angles when he used it himself in his earlier post lol...

Anyway you have the answer, just be mindful that you're actually solving for the angle you're doing the work for.
Firstly, I quoted that the OP claimed that the assignment stated to use law of sines and law of cosines. I was pointing out that the problem didn't need to use the sum of the angles if the problem statement forbids it which is how I understood the OP's response.

An example of why using sines would have been bad.

If you run into solving for an angle using law of sines you would get something that looks like A = arcsin(.5), obviously these are not the numbers from the problem.

lets solve this, throw it in the calculator or what have you, calculator does magic, and.... it tells you A =30. Wait a second, lets try and go the other way and confirm this by doing sin(30)=A. This yields A= .5 just like it should. But check this out, sin(150)=.5

This is why using law of sines can be dangerous, for any angles located between 0 and 180, cosines is better to use.

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