Law of Sines and Cosines Examples

Two sides of a triangle are 8 units and 12 units long. The angle opposite the 8 unit long side is 35°. What is the angle opposite the side that is 12 units long?

A great way to start this (or any other) problem is to figure out what we know. We know that the sun is a mass of incandescent gas, but that doesn't seem relevant here. Instead, let's say that a = 8 and b = 12, while A, the angle opposite a, is 35°. We want to know B.

We know two of the sides and one angle of the triangle, and our unknown variable is another angle. The Law of Cosines has three sides and one angle, so that doesn't fit the problem. What we want is the Law of Sines.

That's the ticket. Let's put our values in there:

Now let's move some things around and get calculating:

We're not done yet, though, we need to apply some inverse sine to both sides to get to B itself.

B = sin^-1 (0.860)

B = 59.4°

What once was unknown is now known.

A triangle has sides measuring 12, 20, and 21 units long. What is the measurement of the angle opposite the side that is 20 units long?

Let's label what we know, what we want to know, and then go from there: a = 12, b = 20, c = 21, B = ?.

With three sides and a missing angle, we need the Law of Cosines, and we need it bad.

b² = a² + c² – 2ac cos B

All aboard the plug-and-chug train. Next stop: cos B.

20² = 12² + 21² – 2(12)(21)cos B

400 = 144 + 441 – 504cos B

400 = 585 – 504cos B

-185 = -504cos B

0.367 = cos B

To find B, all we have left to do is find the inverse:

B = cos^-1 (0.367)

B = 68.5°

A triangle has sides measuring 3 units and 4 units, with an included angle of 57°. What is the length of the side opposite the 57° angle?

Three sides, and one angle? Sounds like a party to us, and the Law of Cosines is invited.

It's pretty easy to label the sides a = 3, b = 4, and c = (who knows). How do we label the angle, 57°? It is an "included" angle, in between a and b; that means it cannot be opposite a or b, so it must be opposite c. That, and we're told that it is opposite our unknown side—who was that again?

Let the solving commence:

c² = a² + b² – 2ab cos C

c² = 3² + 4² – 2(3)(4)cos (57)

c² = 9 + 16 – 24(0.545)

c² = 25 – 13.07

c² = 11.93

c = 3.5 units