Law of Cosines Examples

Find the value of x. 

Law of Cosines, here we come. In this case, we'll say that b = 9 and c = 13. That leaves x = a and A = 42°.

a² = b² + c² – 2bc cos A

Substitute in our values and solve.

x² = 9² + 13² – 2(9)(13) cos(42°)
x² ≈ 76.1
x ≈ 8.7

That's all, folks.

A triangle has side lengths of 10, 11, and 12. Find all the angles.

First, we should assign an angle to each side. A, B, and C will be the angles opposite the sides in increasing order. That means a = 10, b = 11, and c = 12. We'll find A first.

a² = b² + c² – 2bc cos A

To minimize confusion, we'll rearrange the equation before plugging things in.

Wonderful. Now we're ready to plug in our sides.

Our calculator can handle the rest.

A ≈ 51.3°

Noice. Now we have an angle-side ratio, so we'll make use of the Law of Sines for the rest.

We just calculated A and we want to calculate B.

Rearrange the equation to solve for B.

Calculate the angle.

B ≈ 59.1°

Sweet. Two down, one to go.

A + B + C = 180°
51.3° + 59.1° + C = 180°

For the last angle, we get:

C = 69.6°

That means the angles of the triangle are 51.3°, 59.1°, and 69.6°.

Find the value of x, y, and z.

First, we'll start with x, applying the Law of Cosines.

a² = b² + c² – 2bc cos A

If a = x, then b = 121, c = 137, and A = 53°.

x² = 121² + 137² – 2(121)(137)cos(53°)

Solve for x² and then take the square root.

x ≈ 116

To solve for y, we can use the Law of Sines. In our case, y = B.

Substitute in our values.

Solve for B.

B = y ≈ 70.6°

All that's left is z. Now that we've found x, we'll use the Law of Cosines again to calculate z, but this time focusing on the lower triangle.

a² = b² + c² – 2bc cos A

We'll rearrange this equation a bit to solve for the angle.

We need to be careful when substituting in our values. If we set z = A, the side opposite that is a = x ≈ 116. The other two variables, b and c, are 85 and 59.

A = z ≈ 105.9°