SUMMARY OF LAW OF SINES AND LAW OF COSINES

For both the Law of Sines and Law of Cosines, it is simply a matter of deciding which to use and then plugging in the numbers. Interpretation of the answers is fairly simple with the slight exception of the ambiguous case of the Law of Sines. sin A sin B Law of Sines a = b Law of Cosines a2 = b2 + c2 - 2bc cos A

Both formulas can be changed to use different known information.

The known information determines which of the two you should use.

AAA This gives similar triangles but without a side length we can't do anything. ASA In this case we really have AASA, since the sum of the angles of a triangle is 180. AAS Use the Law of Sines. No ambiguity here since for a given AAS there is only one possible triangle. SAS Use the Law of Cosines. SSA Use the Law of Sines. This is the ambiguous case. SSS Use the Law of Cosines.

The above list tells what you would want to use to find the first unknown that you need. The remaining ones can sometimes be found in either of the two ways. In those cases where you are looking for an angle, it is generally best to use the Law of Cosines (even though the Law of Sines is easier to enter into the calculator) because there is no ambiguity.

In all of the following examples, we will refer to the following picture.

Example #1 - If A = 38.4, B = 54.1 and a = 14, find c, C and b.

This gives us AAS, so we will use the Law of Sines. b a a Sin B Sin B = Sin A gives b = Sin A and thus b  18.2574. Subtraction gives C  87.5. Finally, c a a Sin C Sin C = Sin A giving c = Sin A or c  22.5174. This puts the largest side opposite the largest angle and the smallest side opposite the smallest angle.

Example #2 - If a = 1, b = 2 and A = 20, find c, B and C for any possible triangles.

This gives SSA so we will use the Law of Sines. Keep in mind that this is the ambiguous case. We first have Sin B Sin A b sin A b = a which gives Sin B = a . From that we get 2 Sin 20 Sin B = 1 or Sin B  .6840. Now, B = sin-1.6840  43.1601. However, the angle 180 - 43.1601 = 136.8399 gives the same sine value. We therefore have two triangles possible.

Triangle #1 - B  43.1601. Then C  180 - 20 - 43.1601  c a a Sin C 116.8399. Finally, Sin C = Sin A giving c = Sin A or c  2.6088. This does leave the largest side opposite the largest angle and the smallest side opposite the smallest angle so we suspect our work was probably right.

Triangle #2 - B  136.8399. Then C  180 - 136.8399 - 20 c a a Sin C  23.1601. Finally, Sin C = Sin A giving c = Sin A or c  1.1499. This also leaves the largest side opposite the largest angle and the smallest side opposite the smallest angle so we suspect our work was probably right.

Example #3 - If a = 8, b = 5 and A = 31, find c, B and C for any possible triangles. We have SSA so there is ambiguity. We first have Sin B Sin A b sin A b = a which gives Sin B = a . From that we get 5 Sin 31  -1 Sin B = 8 or Sin B  .3218. Sin .3218  18.7718. Another angle between 0 and 180 gives the same sine value. That angle is 180 - 18.7718 or 161.2282, but with A = 31, that is too big an angle. Thus we have only one triangle possible. Subtraction gives C  130.2282. Finally, c a a Sin C Sin C = Sin A giving c = Sin A or c  11.8589. Again, the largest side is opposite the largest angle and the smallest side is opposite the smallest angle so we suspect our answers are correct.

Example #4 - If a = 2, b = 3 and A = 91, find c, B and C for any possible triangles. This is SSA so we use the Law of Sines where we have ambiguity. We first have Sin B Sin A b sin A b = a which gives Sin B = a . From that we get 3 Sin 91 Sin B = 2 or Sin B  1.4997. This is impossible so there are no possible triangles.

Example #5 - If a = 22.1, c = 17.5 and B = 109, find A, C and b. This gives SAS so we use the Law of Cosines. We have b2 = a2 + c2 - 2ac cos B = (22.1)2 + (17.5)2 - 2(22.1)(17.5) cos 109  1046.4869. Thus we have b  32.3494. We then use the Law of Cosines or Law of Sines to find A. In this case, there will be no ambiguity because we know that such a triangle exists and that A must be only less then 90 because B = 109. Using a2 - b2 - c2 the Law of Cosines gives A = Cos-1  or A  Cos-1.7633  -2bc   40.2440. Finally, subtraction gives C  30.7560. Again, we have the smallest side opposite the smallest angle and the largest side opposite the largest angle.

Example #6 - If a = 32.1, b = 27.5 and c = 29.4, find A, B and C. This gives us SSS so we use the Law of Cosines. Using a2 - b2 - c2 the Law of Cosines gives A = Cos-1  or A  -2bc  b2 - a2 - c2  Cos-1.3649  68.5985. Then B = Cos-1  or B  -2ac   Cos-1.6031  52.9077. Finally, subtraction gives us C  180 - 52.9077 - 68.5985  58.4938. The smallest side is opposite the smallest angle and the largest side is opposite the largest angle.