# 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

${d^2} = {a^2} + {a^2} - 2{a^2}\cos 120^\circ = 2{a^2} - 2{a^2} \cdot \left( { - \frac{1}{2}} \right) = 2{a^2} + {a^2} = 3{a^2}.$

Then the least diagonal is equal to

$d = a\sqrt 3 .$

### 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

${c^2} = {a^2} + {b^2} - 2ab\cos 135^\circ .$

Calculate $$\cos 135^\circ$$ using the Reduction Identity:

$\cos 135^\circ = \cos \left( {180^\circ - 45^\circ } \right) = - \cos 45^\circ = - \frac{{\sqrt 2 }}{2}.$

Hence,

${c^2} = {a^2} + {b^2} - 2ab\left( { - \frac{{\sqrt 2 }}{2}} \right) = {a^2} + {b^2} + ab\sqrt 2 , \Rightarrow c = \sqrt {{a^2} + {b^2} + ab\sqrt 2 } .$

### 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:

$\frac{3}{5} = \frac{{{{\left( {x - 1} \right)}^2} + {{\left( {x + 1} \right)}^2} - {x^2}}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}.$

Solve this equation for $$x:$$

$\frac{3}{5} = \frac{{{x^2} - \cancel{{2x}} + 1 + \cancel{{{x^2}}} + \cancel{{2x}} + 1 - \cancel{{{x^2}}}}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}, \Rightarrow \frac{3}{5} = \frac{{{x^2} + 2}}{{2\left( {{x^2} - 1} \right)}}, \Rightarrow 5{x^2} + 10 = 6{x^2} - 6, \Rightarrow {x^2} = 16.$

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:

${3^2} + {4^2} = {5^2}.$

In such a case, the area of the triangle is

$A = \frac{{3 \cdot 4}}{2} = 6.$

### 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 $$ABC:$$

$A{B^2} = A{C^2} + B{C^2} - 2 \cdot AC \cdot BC \cdot \cos 120^\circ .$

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

${7^2} = {6^2} + {x^2} - 2 \cdot 6 \cdot x \cdot \left( { - \frac{1}{2}} \right), \Rightarrow 49 = 36 + {x^2} + 6x, \Rightarrow {x^2} + 6x - 13 = 0.$

Find the discriminant of the quadratic equation:

$D = {6^2} - 4 \cdot 1 \cdot \left( { - 13} \right) = 36 + 52 = 88.$

The roots of the equation are

${x_{1,2}} = \frac{{ - 6 \pm \sqrt {88} }}{2} = \frac{{ - 6 \pm 2\sqrt {22} }}{2} = - 3 \pm \sqrt {22} .$

We choose the positive root, so

$BC = x = - 3 + \sqrt {22} .$

### 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 $$ABC:$$

${c^2} = {a^2} + {b^2} - 2ab\cos \gamma.$

Solve this equation for $$\cos \gamma:$$

$\cos \gamma = \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}}.$

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

$m_a^2 = {\left( {\frac{a}{2}} \right)^2} + {b^2} - 2 \cdot \frac{a}{2} \cdot b\cos \gamma = \frac{{{a^2}}}{4} + {b^2} - ab\cos \gamma .$

Subctitute $$\cos \gamma$$ found above:

$m_a^2 = \frac{{{a^2}}}{4} + {b^2} - ab \cdot \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}} = \frac{{{a^2}}}{4} + {b^2} - \frac{{{a^2} + {b^2} - {c^2}}}{2} = \frac{{{a^2} + 4{b^2} - 2{a^2} - 2{b^2} + 2{c^2}}}{4} = \frac{{2{c^2} + 2{b^2} - {a^2}}}{4} = \frac{1}{4}\left( {2{c^2} + 2{b^2} - {a^2}} \right).$

### 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.

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

$\angle B = 180^\circ - \alpha .$

Applying the Law of Cosines to the triangle $$ABC,$$ we can write:

$A{C^2} = A{B^2} + B{C^2} - 2 \cdot AB \cdot BC \cdot \cos \left( {180^\circ - \alpha } \right).$

By the Reduction Identity,

$\cos \left( {180^\circ - \alpha } \right) = - \cos \alpha .$

This yields:

$d_1^2 = {a^2} + {b^2} - 2ab\left( { - \cos \alpha } \right) = {a^2} + {b^2} + 2ab\cos \alpha .$

Similarly we find the square of the diagonal $$d_2:$$

$B{D^2} = A{B^2} + A{D^2} - 2 \cdot AB \cdot AD \cdot \cos \alpha ,$

or

$d_2^2 = {a^2} + {b^2} - 2ab\cos \alpha .$

Calculate the sum of the squares of the diagonals:

$d_1^2 + d_2^2 = {a^2} + {b^2} + \cancel{{2ab\cos \alpha }} + {a^2} + {b^2} - \cancel{{2ab\cos \alpha }} = 2\left( {{a^2} + {b^2}} \right).$

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