# An SAS triangle can't have two solutions... but this one is strange

• Apr 20th 2008, 01:45 PM
Mpromptu
An SAS triangle can't have two solutions... but this one is strange
<A = 34 degrees. b=24. c=46.

This is an SAS triangle... when I use Law of Cosines, I solved for side a and got a=29.35.

Then I tried using Law of Sines to solve for <B and <C.

If I solve for <B first, I set it up this way:

sin(34)
-------
29.35

=

sin B
--------
24

and I get <B=27.2 degrees. Which means <C=118.8 degrees.

However, if I solve for <C first, I set it up this way:

sin(34)
-------
29.35

=

sin C
--------
46

and I get <C=61.2 degrees. Which means <B=84.8 degrees.

What is going on?! I know I'm missing something here, but I can't figure out what it is!

• Apr 21st 2008, 12:12 AM
earboth
Quote:

Originally Posted by Mpromptu
<A = 34 degrees. b=24. c=46.

a=29.35. OK

Then I tried using Law of Sines to solve for <B and <C.

<C=118.8 degrees. OK

However, if I solve for <C first, I set it up this way:

sin(34)
-------
29.35

=

sin C
--------
46

and I get <C=61.2 degrees.

1. If you consider the lengths of the sides of your triangle you'll see that the angle C must be very much larger than the other two angles.

2. From

$\frac{\sin(34^\circ)}{29.35} = \frac{\sin(C)}{46} ~\implies~\sin(C)=0.876418...$

And with this value you get two different angles for C:

$C\approx 61.2^\circ~\vee~ C=180^\circ-61.2^\circ=118.8^\circ$
• Apr 21st 2008, 02:04 PM
Mpromptu
Thanks for the reply, but I don't think my question has been answered.

Obviously, <C must be the largest of my two choices; it's opposite the largest side. Choosing the smaller <C would make no sense, visually.

My question is why I have two choices for <C to begin with! This is an SAS triangle. Shouldn't there only be one option for my missing information?

For what reasons can I dismiss the smaller C? Is there any mathematical reason I'm even coming up with that answer in the first place?
• Apr 22nd 2008, 10:42 AM
earboth
Quote:

Originally Posted by Mpromptu
Thanks for the reply, but I don't think my question has been answered.

Obviously, <C must be the largest of my two choices; it's opposite the largest side. Choosing the smaller <C would make no sense, visually.

My question is why I have two choices for <C to begin with! This is an SAS triangle. Shouldn't there only be one option for my missing information?

For what reasons can I dismiss the smaller C? Is there any mathematical reason I'm even coming up with that answer in the first place?

Let us pretend $\angle C = 61.2^\circ$

With $\angle A = 34^\circ$ you'll get $\angle B = 84.8^\circ$

That means $\angle B$ is the greatest angle but b is the smallest side, which is impossible. Therefore $\angle C = 61.2^\circ$ must be false.
• Apr 23rd 2008, 07:00 PM
thetreece
It's because your calculator is picking it up in the first quadrant rather than the second. If you remember your good ole unit circle, you understand the sine of say. . . 60 degrees will be the same as the sine of 120 degrees. When you take that inverse sine in the end it is giving you a right answer, just in the wrong quadrant.
Check this out: 90-61.2=28.8
118.8-90=28.8

Do you dig that? I know the best way to avoid this with an SSS triangle is to always solve for the biggest angle first. I hope you sort of understand now and can get around this in the future.
• Apr 23rd 2008, 07:12 PM
Mpromptu
Thanks, thetreece
I think that's the issue. I appreciate the help!