# Law of Cosines

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the least diagonal of a regular hexagon with side length *a*.

### Example 2

Two sides of a triangle are equal to *a* and *b*. The angle between them is 135°. Find the third side *c*.

### Example 3

The middle side of a triangle is \(1\) greater than the least side and \(1\) less than the greatest side. The cosine of the middle angle is equal to \(\frac{3}{5}.\) Find the area of the triangle.

### Example 4

In a triangle \(ABC,\) the side \(AB\) is \(7,\) the midline parallel to the side \(AC\) is \(3,\) and the angle \(\angle C = 120^\circ.\) Find the side \(BC.\)

### Example 5

Given a triangle whose sides are \(a, b,\) and \(c;\) \(m_a\) is the length of the median drawn from the vertex \(A\) to side \(a.\) Prove that \[m_a^2 = \frac{1}{4}\left( {2{b^2} + 2{c^2} - {a^2}} \right).\]

### Example 6

Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

### Example 1.

Find the least diagonal of a regular hexagon with side length \(a.\)

Solution.

The internal angle in a regular hexagon is \(120^\circ.\) Using the Cosine Rule, we get

Then the least diagonal is equal to

### Example 2.

Two sides of a triangle are equal to \(a\) and \(b.\) The angle between them is \(135^\circ.\) Find the third side \(c.\)

Solution.

By the Cosine Rule, we have

Calculate \(\cos 135^\circ\) using the Reduction Identity:

Hence,

### Example 3.

The middle side of a triangle is \(1\) greater than the least side and \(1\) less than the greatest side. The cosine of the middle angle is equal to \(\frac{3}{5}.\) Find the area of the triangle.

Solution.

Let \(x\) be the length of the middle side. Then the other sides of the triangles are equal to \(x - 1\) and \(x + 1.\)

In any triangle, the mid-sized side and mid-sized angle are opposite to each other. Hence, using the Law fo Cosines, we can write:

Solve this equation for \(x:\)

The negative root \(x = -4\) does not make sense. Hence, \(x = 4.\) The other two sides are equal \(x - 1 = 3\) and \(x + 1 = 5.\)

Note that this triangle satisfies the Pythagorean theorem:

In such a case, the area of the triangle is

### Example 4.

In a triangle \(ABC,\) the side \(AB\) is \(7,\) the midline parallel to the side \(AC\) is \(3,\) and the angle \(\angle C = 120^\circ.\) Find the side \(BC.\)

Solution.

The midline \(KM\) is half the length of the side \(AC,\) that is, \(AC = 6.\)

Let \(x\) be the length of the side \(BC.\) Write the Cosine Rule for the triangle \(ABC:\)

Recall that \(\cos 120^\circ = - \cos 60^\circ = - \frac{1}{2}.\) Then

Find the discriminant of the quadratic equation:

The roots of the equation are

We choose the positive root, so

### Example 5.

Given a triangle whose sides are \(a, b,\) and \(c;\) \(m_a\) is the length of the median drawn from the vertex \(A\) to side \(a.\) Prove that \[m_a^2 = \frac{1}{4}\left( {2{b^2} + 2{c^2} - {a^2}} \right).\]

Solution.

Use the Law of Cosines for the triangle \(ABC:\)

Solve this equation for \(\cos \gamma:\)

Apply now the Law of Cosines to the triangle \(AMC\) taking into account that \(CM = \frac{{BC}}{2} = \frac{a}{2}.\) This yields:

Subctitute \(\cos \gamma\) found above:

### Example 6.

Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Solution.

Consider a parallelogram \(ABCD\) with sides \(a, b\) and diagonals \(d_1,\) \(d_2.\)

Denote the angle \(A\) by \(\alpha.\) The other angle \(B\) is then equal to

Applying the Law of Cosines to the triangle \(ABC,\) we can write:

By the Reduction Identity,

This yields:

Similarly we find the square of the diagonal \(d_2:\)

or

Calculate the sum of the squares of the diagonals:

The right-hand side represents the sum of the squares of four sides of the parallelogram.