# SAS with Laws of Sine and Cosine

• Dec 5th 2010, 11:35 AM
JennyFlowers
SAS with Laws of Sine and Cosine
Hello. I had to solve the following triangle using Law of Sines / Cosines:

b = 6
c = 13
alpha = 16 degrees

The first time I solved the triangle, I first used the law of cosines and got

a = 7.42

I then used the law of sines to solve for beta and got 12.9 degrees. (beta = sin inverse of 6 sin 16 / 7.42)

I then used 180 - (16 + 12.9) and found gamma to be 151.1 degrees.

I then solved the triangle a second time, but instead of solving for the beta angle using the law of sines, I solved for gamma.

In that case, I found that gamma was equal to 28.8 degrees. (gamma = sin inverse of 13 sin 16 / 7.42) This would then make beta equal to 135.2 degrees.

I read that I'm supposed to solve for the smaller angle first using the law of sines. Can solving for the larger angle first cause such a wide variation in the answer, or did I incorrectly solve for the larger angle?

Thanks!
• Dec 5th 2010, 11:51 AM
pickslides
I get

$\displaystyle a = \sqrt{6^2+13^2-2\times 6\times 13\times \cos{16}}\approx 18.83$

Now you can use the sine rule as you have described.
• Dec 5th 2010, 12:02 PM
JennyFlowers
Quote:

Originally Posted by pickslides
I get

$\displaystyle a = \sqrt{6^2+13^2-2\times 6\times 13\times \cos{16}}\approx 18.83$

Now you can use the sine rule as you have described.

Your answer is the result of solving in radian mode. This problem provides values in degrees.

My solution for 'a' should be correct.
• Dec 5th 2010, 12:08 PM
pickslides
True, my apologies for that, it is still rather early here.

The only variataion you should experience is though rounding deciamls in previous steps. Otherwise your answers should be relatively close.
• Dec 5th 2010, 12:11 PM
JennyFlowers
The answers are not close at all, unfortunately. Can you see where I went wrong?

Shouldn't gamma = sin inverse of 13 sin 16 / 7.42?
• Dec 5th 2010, 12:19 PM
pickslides
Quote:

Originally Posted by JennyFlowers
Shouldn't gamma = sin inverse of 13 sin 16 / 7.42?

Makes sense to me. Maybe the answer sheet is wrong, it happens...
• Dec 5th 2010, 12:26 PM
JennyFlowers
I was able to replicate this with another problem.

SAS with:

a = 9
c = 12
beta = 23 degrees

Using law of cosines I find:

b = 5.12

Using law of sines to solve for alpha, I find:

alpha = sin inverse of 9 sin 23 / 5.12 = 43.38 degrees

That would mean that gamma = 180 - (23 + 43.38) = 113.62 degrees

According to my text book, this is the correct solution, and I found it by using law of sines on the smaller angle.

Now, when I use law of sines on the larger angle, I get:

gamma = sin inverse of 12 sin 23 / 5.12 = 66.31 degrees.

So solving it the first way I got gamma at 43.38 degrees and the second way, 66.31 degrees. Huge variation!

It seems that it is necessary to solve this type of triangle with the small angle first when using law of sines.

Can someone either confirm or disprove this?

Thanks!
• Dec 5th 2010, 12:30 PM
JennyFlowers
Quote:

Originally Posted by pickslides
Makes sense to me. Maybe the answer sheet is wrong, it happens...

There was no answer sheet for that particular problem. What I don't understand is why my two different solutions don't agree. As you suggested, they should be very close, but they are not. Can you confirm this by working the problem?
• Dec 5th 2010, 12:52 PM
Ylani
Actually you mean the first way $\gamma$ is 113.62, and the second way $\gamma$ is 66.31. The problem is that angles in the second quadrant share sine values with angles in the first quadrant. The inverse sine function is set to only give you angles from -90 degrees to + 90 degrees. Therefore you will not be able to get an answer of 113.62 using the law of sines with the inverse sine function. You get an angle that has the desired sine value, but it is not the right one.

To avoid this problem, look in advance for the largest angle in the triangle. The largest angle is always opposite the longest side. Then solve for the smaller angles first.

Hope that helps.
• Dec 5th 2010, 12:55 PM
JennyFlowers
Thank you, that's exactly what I was looking for. I'm glad I stumbled upon this exception today, finals are tomorrow!
• Dec 5th 2010, 01:10 PM
Ylani
Good work on your part trying something two different ways. Good luck on your finals!