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## Law of Cosines

The **Law of Cosines** is used to find the remaining parts of an oblique (non-right) triangle when either the lengths of two sides and the measure of the included angle is known (SAS) or the lengths of the three sides (SSS) are known. In either of these cases, it is impossible to use the Law of Sines because we cannot set up a solvable proportion.

The Law of Cosines states:

*c ^{2}=a^{2}+b^{2}−2ab cosC .*

This resembles the Pythagorean Theorem except for the third term and if *C* is a right angle the third term equals 0 because the cosine of 90° is 0

and we get the Pythagorean Theorem. So, the Pythagorean Theorem is a special case of the Law of Cosines.

The Law of Cosines can also be stated as

Since cos*B* is negative, we know that *B*

is an obtuse angle.

*B*≈116.80°

Since *B* is an obtuse angle and a triangle has at most one obtuse angle, we know that angle *A* and angle *C*

are both acute.

To find the other two angles, it is simplest to use the Law of Sines.

## When to Use

The Law of Cosines is useful for finding:

- the third side of a triangle when we know
**two sides and the angle between**them (like the example above) - the angles of a triangle when we know
**all three sides**(as in the following example)

But it is easier to remember the “**c ^{2}**=” form and change the letters as needed !

As in this example:

Answer: z = 14.5

Did you notice that cos(131º) is negative and this changes the last sign in the calculation to + (plus)? The cosine of an obtuse angle is always negative (see Unit Circle).

This law can be derived in a number of ways. The definition of the dot product incorporates the law of cosines, so that the length of the

## The Laws of Cosines and Sines

We saw in the section on oblique triangles that the law of cosines and the law of sines were useful in solving for parts of a triangle if certain other parts are known. The question here is “why are those laws valid?”

This is an optional section. When learning how to use trigonometry to solve oblique triangles, it is most important to know when and how to use these two laws. If that’s enough for you, then just skip on to the next section on area of a triangle. But if you’re interested in why they’re true, then continue on. As usual, we’ll use a standard notation for the angles and sides of a triangle. That means the side a is opposite the angle A, the side b is opposite the angle B, and the side c is opposite the angle C. |

First, drop a perpendicular line *AD* from *A* down to the base *BC* of the triangle. The foot *D* of this perpendicular will lie on the edge *BC* of the triangle when both angles *B* and *C* are acute. But if angle *B* is obtuse, then the foot *D* will lie on *BC* extended in the direction of *B.* Yet if angle *C* is obtuse, then *D* will line on *BC* extended in the direction of *C.* Fortunately, the argument is the same in all three cases.

Let *h* denote the length of this line *AD,* that is, the height (or altitude) of the triangle.

When angle *B* is acute, then sin *B* = *h/c.* But this is true even when *B* is an obtuse angle as in the third diagram. There, angle *ABC* is obtuse. But the sine of an obtuse angle is the same as the sine of its supplement. That means sin *ABC* is the same as sin *ABD,* that is, they both equal *h/c.*

Likewise, it doesn’t matter whether angle *C* is acute or obtuse, sin *C* = *h/b* in any case.

These two equations tell us that *h* equals both *c* sin *B* and *b* sin *C.* But from the equation *c* sin *B* = *b* sin *C,* we can easily get the law of sines:

Since the three verions differ only in the labelling of the triangle, it is enough to verify one just one of them. We’ll consider the version stated first.

In order to see why these laws are valid, we’ll have to look at three cases. For case 1, we’ll take the angle *C* to be obtuse. In case 2, angle *C* will be a right angle. In case 3, angle *C* will be acute.

Case 1. For this case, we take angle C to be obtuse. This case has a wrinkle in it since the cosine of an obtuse angle is negative. Let’s see how that goes. |

First, drop a perpendicular line AD from A down to the base BC of the triangle. In this case, the foot D of this perpendicular will lie outside the triangle. Let h denote the height of the triangle, let d denote BD, and let e denote CD. We can derive the following equations from the figure: |

In general, the cosine of an obtuse angle is the negation of the cosine of its supplement. In this case that means the cosine of angle *C,* that is to say angle *ACB,* is the negation of the cosine of angle *ACD.* That’s why the minus sign appears in the last equation.

These equations and plain algebra finish the argument as follows:

Thus, the law of cosines is valid when *C* is an obtuse angle.

**Case 2.** Now consider the case when the angle at *C* is right. The cosine of a right angle is 0, so the law of cosines, *c*^{2} = *a*^{2} + *b*^{2} – 2*ab* cos *C,* simplifies to becomes the Pythagorean identity, *c*^{2} = *a*^{2} + *b*^{2}, for right triangles which we know is valid.

**Case 3.** In this case we assume that the angle *C* is an acute triangle. Drop a perpendicular line *AD* from *A* down to the base *BC* of the triangle. The foot *D* of the perpendicular will (1) lie on the edge *BC* if angle *B* is acute, (2) coincide with the point *B* if the angle *B* is right, or (3) lie on the side *BC* extended if the angle *B* is obtuse.

Let *h* denote the height of the triangle, let *d* denote *BD,* and *e* denote *CD.*

Then we can read the following relationships from the diagram:

c^{2} | = | d^{2} + h^{2} |

b^{2} | = | e^{2} + h^{2} |

cos C | = | e/b |

d^{2} | = | (e — a)^{2} |

That last equation requires explanation. If the point *D* lies on the side *BC,* then *d* = *a* – *e,* but if *D* lies on *BC* extended, then *d* = *e* – *a.* In either case, *d*^{2} = (*e* – *a*)^{2}.

These equations and a little algebra finish the proof as follows:

c^{2} | = | d^{2} + h^{2} |

= | d^{2} – e^{2} + b^{2} | |

= | (d – e) (d + e) + b^{2} | |

= | (a – 2e) a + b^{2} | |

= | a^{2} + b^{2} – 2ae | |

= | a^{2} + b^{2} – 2ab cos C |

Thus, we now know that the law of cosines is valid when both angle *C* is acute, and we’ve finished all three cases.

Incidentally, Euclid included in his *Elements* a couple of propositions, II.12 and II.13, that look very much like the law of cosines, but they are not actually the law of cosines, of course, since trigonometry had not been developed in Euclid’s time.

## Law of Cosines

The law of cosines relates the lengths of the sides of a triangle to the cosine of one of its angles. Using trigonometry, we can now obtain values of distances and angles which cannot be measured otherwise. The law of cosines finds application while computing the third side of a triangle given two sides and their enclosed angle, and for computing the angles of a triangle if all three sides are known to us.

A triangle has 6 elements (3 sides + 3 angles). Let us understand the law of the cosines formula and its derivation to study the inter-relationship of these elements using the cosine function.

## What is Law of Cosines?

The law of cosine helps in establishing a relationship between the lengths of sides of a triangle and the cosine of its angles. The cosine law in trigonometry generalizes the Pythagoras theorem, which applies to a right triangle.

### Law of Cosines: Definition

**Statement:** The law of cosine states that the square of any one side of a triangle is equal to the difference between the sum of squares of the other two sides and double the product of other sides and cosine angle included between them.

Let a, b, and c be the lengths of the three sides of a triangle and A, B, and C be the three angles of the triangle. Then, the law of cosine states that: a^{2} = b^{2} + c^{2} − 2bc·cosA. As stated above, the law of cosines in trigonometry generalizes the Pythagorean theorem. If you plug 90º for the angle in one of the rules, what will happen? Since cos 90º = 0, we are left with the Pythagoras theorem.

The law of cosine is also known as the cosine rule. This law is useful to find the missing information in any triangle. For example, if you know the lengths of two sides of a triangle and an angle included between them, this rule helps to find the third side of the triangle. Let us check out different cosine law formulas and the method to find these missing parameters in the following sections.

## Law of Cosines Formula

The law of cosines formula can be used to find the missing side of a triangle when its two sides and the included angle is given i.e., it is used in the case of a SAS triangle. We know that if A, B, and C are the vertices of a triangle, then their opposite sides are represented by the small letters a, b, and c respectively. The law of cosines formula is used to:

- find a when b, c, and A are given (or)
- find b when a, c, and B are given (or)
- find c when a, b, and C are given (or)
- find any angle of the triangle when a, b, and c are given.

There are three laws of cosines and we choose one of them to solve our problems depending on the available data.

a^{2} = b^{2} + c^{2 }– 2bc·cosA

b^{2} = c^{2} + a^{2} – 2ca·cosB

c^{2} = a^{2} + b^{2} – 2ab·cosC

## Proof of Law of Cosines

There is more than one way to prove the law of cosine. Let’s prove it using trigonometry. Consider the following figure.

In ΔABM we have,

sin A = BM/AB = h/c

and,

cos A = AM/AB = r/c

From equation (1) and (2), we get h = c(sin A) and r = c(cos A)

By Pythagoras Theorem in ΔBMC,

a^{2 }= h^{2} + (b – r)^{2}

Substitute h = c(sin A) and r = c(cos A) in equation (3).

a^{2} = (c(sinA))^{2} + (b – c(cosA))^{2}

= c^{2}sin^{2}A + b^{2} + c^{2}cos^{2}A – 2bc·cosA

= c^{2}(sin^{2}A + cos^{2}A) + b^{2} – 2bc·cosA

= c^{2} + b^{2} – 2bc·cosA

Hence, proved.

## Finding Missing Length and Angles Using the Law of Cosines

As discussed above, law of cosines can be used to calculate the missing parameters of a triangle, given the required known elements. Let us have a look at the following steps to understand the process of finding the missing side or angle of a triangle using the cosine law.

**Step 1:**Note down the given data(side lengths and measure of angles) for the triangle and identify the element to be calculated.**Step 2:**Apply the cosine rule formulas,

a^{2}= b^{2}+ c^{2 }– 2bc·cosA

b^{2}= c^{2}+ a^{2}– 2ca·cosB

c^{2}= a^{2}+ b^{2}– 2ab·cosC

where, A, B, and C are the vertices of a triangle, and their opposite sides are represented as a, b, and c respectively.**Step 3:**Express the obtained result with suitable units.

Let us consider a few examples to find the missing side and angle of a triangle.

**Example:** Look at the figure shown below.

We need to find the measure of ∠A.

We will use the formula

a^{2} = b^{2} + c^{2} – 2bc.cosA. Substitute 10 for ‘a’, 7 for ‘b’ and 5 for ‘c’.

10^{2 }= (7)^{2} + (5)^{2} – 2(5)(7)·cosA

70·cos A = -26

cos A = -13/35

A = 111.8º

In this example, we used the law of the cosine equation to find the missing angle. Now, let us use the law of the cosine equation to find the missing side.

**Example:** Two sides of a triangle measure 72 in and 50 in with the angle between them measuring 49º let us find the missing side.

**Solution:**

Substitute 72 for b, 50 for c and 49º for A.

Using the law of cosines formula,

a^{2 }= b^{2} + c^{2} – 2bc·cosA

a^{2} = (72)^{2} + (50)^{2} – 2(72)(50)cos49º

a^{2} = 5184 + 2500 – (7200)(0.656)

a^{2} = 5184 + 2500 – 4723.2

a^{2} = 2960.8

a ≈ 54.4

So, the missing length of the side is 54.4 inches.

**Important Notes on Law of Cosines:**

- Three different versions of the law of cosine are: a
^{2}= b^{2}+ c^{2 }– 2bc·cosA

b^{2}= c^{2}+ a^{2}– 2ca·cosB

c^{2}= a^{2}+ b^{2}– 2ab·cosC - Pythagoras Theorem is a generalization of the Law of Cosine.
- The law of cosine can be applied in any triangle.

**Challenging Question: A spider is lost in its web. Look at the figure below. Can you find the value of x?**

## Examples Using Law of Cosines

**Example 1: A boy is standing at point A and two boats are located at points, B and C, such that the positions of all three form a triangle. If the measure of angle A is 36º with the lengths AB and AC measuring 2.5 ft and 1.8 ft respectively, determine the distance between the two boats floating at the lake.**

**Solution:**

The law of cosine is expressed in three different ways.

We will use the formula a^{2} = b^{2} + c^{2} – 2bc·cos A, because the required side is opposite to ∠A.

Substitute 1.8 for ‘b’, 2.5 for ‘c’ and 36º for angle A.

a^{2} = b^{2} + c^{2} – 2bc·cosA

a^{2} = (1.8)^{2} + (2.5)^{2} – 2(1.8)(2.5)cos 36º

a^{2} = 3.24 + 6.25 – (9)(0.8)

a^{2} = 3.24 + 6.25 – 7.2

a^{2} = 2.29

a ≈ 1.5

**Answer: The length across the boats is 1.5 feet.**

**Example 2: A farmer has a huge field in the shape of a triangle. The two sides of the field measure 624 ft and 327 ft and the angle between them measures 93º. Calculate how much fencing is needed to enclose the field?**

**Solution:**

We will use the definition of law of cosine: a^{2} = b^{2} + c^{2} – 2bc·cosA.

Substitute 624 for ‘b’, 327 for ‘c’ and 93º for A.

a^{2} = b^{2} + c^{2} – 2bc·cosA

a^{2} = (624)^{2} + (327)^{2} – 2(624)(327)cos 93º

a^{2} = 389376 + 106929 – (408096)(-0.05)

a^{2} = 389376 + 106929 + 20404.8

a^{2} = 516709.8

a ≈ 719

So, the perimeter of the triangular field is 624 + 327 + 719 = 1670ft

**Answer: The farmer will need 1670 ft of fencing.**

**Example 3: The adjacent sides of a parallelogram measure 6 in and 10 in with the angle between them measuring 79º. Can you determine the length of the diagonal of the parallelogram?**

**Solution:**

Given: Let ABCD be a parallelogram, such that, CD = 6 in, BC = 10 in.

We know that the adjacent angles of a parallelogram are supplementary.

So, ∠B = 180º – ∠C = 180º – 79º = 101º.

Also, the opposite sides of a parallelogram are equal.

So, AB = CD = 6 in.

Let’s apply the cosine rule in ΔABC.

Substitute 10 for ‘a’, 6 for ‘c’ and 101º for B.

b^{2} = a^{2} + c^{2} – 2ac·cosB

b^{2} = (10)^{2} + (6)^{2} – 2(10)(6)cos101º

b^{2} = 100 + 36 – (120)(-0.19)

b^{2} = 100 + 36 + 22.8

b^{2} = 158.8

b ≈ 12.6

**Answer: The length of the diagonal is approximately 12.6 inches.**

## FAQs on Law of Cosines

### What is Law of Cosines in Trigonometry?

The law of cosines is used to find the missing side of a triangle when its two sides and the included angle is given. There are three laws of cosines and we choose one of them to solve our problems depending on the available data.

- a
^{2}= b^{2}+ c^{2 }– 2bc·cosA - b
^{2}= c^{2}+ a^{2}– 2ca·cosB - c
^{2}= a^{2}+ b^{2}– 2ab·cosC

where, A, B, and C are the vertices of a triangle, and their opposite sides are represented by the small letters a, b, and c respectively.

### What is Law of Cosines Used For?

The Law of Cosines can be used to find the unknown parts of an oblique triangle(non-right triangle), such that either the lengths of two sides and the measure of the included angle is known (SAS) or the lengths of the three sides (SSS) are given.

### What are the Possible Criteria for Law of Cosines?

In order to use the law of cosines, either two sides of the triangle and the measure of included angle(SAS) or the length of all three sides of the triangle(SSS) should be known.

### What is the Formula for Law of Cosines?

The formula for law of cosines is given as,

- a
^{2}= b^{2}+ c^{2 }– 2bc·cosA - b
^{2}= c^{2}+ a^{2}– 2ca·cosB - c
^{2}= a^{2}+ b^{2}– 2ab·cosC

where, A, B, and C are the vertices of a triangle, and their opposite sides are represented as a, b, and c respectively.

### How to Derive Law of Cosines Formula?

There is more than one way to derive the law of the cosine formula. A few of them are given as,

- Using basic concepts of trigonometry.
- Using vector algebra.
- Using the law of sine.
- Using the coordinate geometry to find the distance between two coordinate points.

### What is the Application of Law of Cosines Formula?

The law of cosines formula finds application in finding the missing side of a triangle when its two sides and the included angle is given i.e., it is used in the case of a SAS triangle.

### How to Find the Missing Side or Angle of Triangle Using Law of Cosines?

Law of cosines can be used to find the missing side or angle of a triangle by applying any of the following formulas,

- a
^{2}= b^{2}+ c^{2 }– 2bc·cosA - b
^{2}= c^{2}+ a^{2}– 2ca·cosB - c
^{2}= a^{2}+ b^{2}– 2ab·cosC

Here, A, B, and C are the vertices of a triangle, and their opposite sides are represented using letters a, b, and c respectively.

### Does the Law of Cosine Work in All Triangles?

Yes, the law of cosines can be applied to all the triangles. It stands true for both right and oblique triangles.

### Who Invented the Law of Cosines?

The elements by Euclid contributed to the discovery of the law of cosines. Jamshīd al-Kāshī, a Persian mathematician, was the first to provide the first explicit statement of the law of cosines in the 15th century.

Note that we did not prove the Law of Cosines for right triangles, since it turns out (see Exercise 15) that all three formulas reduce to the Pythagorean Theorem for that case. The Law of Cosines can be viewed as a generalization of the Pythagorean Theorem.

Also, notice that it suffices to remember just one of the three Equations 2.2.1-2.2.3, since the other two can be obtained by “cycling” through the letters *a*, *b*, and *c*. That is, replace *a* by *b*, replace *b* by *c*, and replace *c* by *a*

(likewise for the capital letters). One cycle will give you the second formula, and another cycle will give you the third.

The angle between two sides of a triangle is often called the **icluded angle**. Notice in the Law of Cosines that if two sides and their included angle are known (e.g. *b*, *c*, and *A*), then we have a formula for the square of the third side. We will now solve the triangle from Example 2.2.2

.Example 2.2.2: Case 3 – Two sides and the angle between them Known

Notice in Example 2.2.2

that there was only one solution. For Case 3 this will *always* be true: when given two sides and their included angle, the triangle will have exactly one solution. The reason is simple: when joining two line segments at a common vertex to form an angle, there is exactly one way to connect their free endpoints with a third line segment, regardless of the size of the angle.

You may be wondering why we used the Law of Cosines a second time in Example 2.2.2, to find the angle *B*

. Why not use the Law of Sines, which has a simpler formula? The reason is that using the cosine function eliminates any ambiguity: if the cosine is positive then the angle is acute, and if the cosine is negative then the angle is obtuse. This is in contrast to using the sine function; as we saw in Section 2.1, both an acute angle and its obtuse supplement have the same positive sine.

To see this, suppose that we had used the Law of Sines to find *B* in Example 2.2.2:

How would we know which answer is correct? We could not immediately rule out *B*=127.5^{∘} as too large, since it would make *A*+*B*=157.5^{∘}<180^{∘} and so *C*=22.5^{∘}, which seems like it could be a valid solution. However, this solution is impossible. Why? Because the largest side in the triangle is *c*=5, which (as we learned in Section 2.1) means that *C* has to be the largest angle. But *C*=22.5^{∘} would not be the largest angle in this solution, and hence we have a contradiction.

It remains to solve a triangle in Case 4, i.e. given three sides. We will now see how to use the Law of Cosines for that case.

Example 2.2.3: Case 4 – Three sides Known

We could have saved ourselves some effort by recognizing that the length of one of the sides (*c*=6) is greater than the sums of the lengths of the remaining sides (*a*=2 and *b*=3), which (as the picture below shows) is impossible in a triangle.

The Law of Cosines can also be used to solve triangles in Case 2 (two sides and one opposite angle), though it is less commonly used for that purpose than the Law of Sines. The following example gives an idea of how to do this.

Example 2.2.5

: Case 2 – Two sides and one opposite angle Known

Solve the triangle △*ABC* given *a*=18, *A*=25^{∘}, and *b*=30.**Solution**

In Example 2.2 from Section 2.1 we used the Law of Sines to show that there are two sets of solutions for this triangle: *B*=44.8^{∘}, *C*=110.2^{∘}, *c*=40 and *B*=135.2^{∘}, *C*=19.8^{∘}, *c*=14.4. To solve this using the Law of Cosines, first find *c* by using the formula for *a*^{2}:

which is a quadratic equation in *c*, so we know that it can have either zero, one, or two real roots (corresponding to the number of solutions in Case 2). By the quadratic formula, we have

which is close to what we found before (the small difference being due to different rounding). The other solution set can be obtained similarly.

Like the Law of Sines, the Law of Cosines can be used to prove some geometric facts, as in the following example.

Example 2.2.6

: Parallelogram Diagonals

Use the Law of Cosines to prove that the sum of the squares of the diagonals of any parallelogram equals the sum of the squares of the sides.

**Solution:**

Let *a* and *b* be the lengths of the sides, and let the diagonals opposite the angles *C* and *D* have lengths *c* and *d*, respectively, as in Figure 2.2.2

. Then we need to show that

## Law of Cosines

**Law of cosines** signifies the relation between the lengths of sides of a triangle with respect to the cosine of its angle. It is also called the cosine rule. If ABC is a triangle, then as per the statement of cosine law, we have:

a^{2} = b^{2} + c^{2} – 2bc cos α, where a,b, and c are the sides of triangle and α is the angle between sides b and c.

Similarly, if β and γ are the angles between sides ca and ab, respectively, then according to the law of cosine, we have:

b^{2} = a^{2} + c^{2} – 2ac cos β

c^{2} = b^{2} + a^{2} – 2ab cos γ

Fact: If any one of the angles, α, β or γ is equal to 90 degrees, then the above expression will justify the Pythagoras theorem, because cos 90 = 0. Hence, the above three equations can be expressed as:

a^{2} = b^{2} + c^{2} [if α = 90 degrees]

b^{2} = a^{2} + c^{2} [if β = 90 degrees]

c^{2} = b^{2} + a^{2} [if γ = 90 degrees]

## Law of Cosines Definition

In Trigonometry, the law of Cosines, also known as Cosine Rule or Cosine Formula basically relates the length of the triangle to the cosines of one of its angles. It states that, if the length of two sides and the angle between them is known for a triangle, then we can determine the length of the third side. It is given by:

c^{2} = a^{2} + b^{2} – 2ab cosγ

Where a, b and c are the sides of a triangle and γ is the angle between a and b. See the figure below.

## Formulas

As per the cosines law formula, to find the length of sides of triangle say △ABC, we can write as;

- a
^{2}= b^{2}+ c^{2}– 2bc cos α - b
^{2}= a^{2}+ c^{2}– 2ac cos β - c
^{2}= b^{2}+ a^{2}– 2ba cos γ

And if we want to find the angles of △ABC, then the cosine rule is applied as;

- cos α = [b
^{2}+ c^{2}– a^{2}]/2bc - cos β = [a
^{2}+ c^{2}– b^{2}]/2ac - cos γ = [b
^{2}+ a^{2}– c^{2}]/2ab

Where a, b and c are the lengths of sides of a triangle.

### Solving SSS Congruency

In SSS congruence, we know the lengths of all the three sides of a triangle, and we need to find the measure of the unknown triangle. Therefore, using the law of cosines, we can find the missing angle.

First we need to find one angle using cosine law, say cos α = [b^{2} + c^{2} – a^{2}]/2bc.

Then we will find the second angle again using the same law, cos β = [a^{2} + c^{2} – b^{2}]/2ac

Now the third angle you can simply find using angle sum property of triangle. That means the sum of all the three angles of a triangle is equal to 180 degrees.

## Proof

Now let us learn the **law of cosines proof** here;

In the right triangle BCD, by the definition of cosine function:

cos C = CD/a

or

CD=a cos C

Subtracting above equation from side b, we get

DA = b − acosC ……(1)

In the triangle BCD, according to Sine definition

sin C = BD/a

or

BD = a sinC ……(2)

In the triangle ADB, if we apply the Pythagorean Theorem, then

c^{2} = BD^{2} + DA^{2}

Substituting for BD and DA from equations (1) and (2)

c^{2} = (a sin C)^{2} + (b-acosC)^{2}

By Cross Multiplication we get:

c^{2} = a^{2} sin^{2}C + b^{2} – 2abcosC + a^{2} cos^{2}C

Rearranging the above equation:

c^{2} = a^{2} sin^{2}C + a^{2} cos^{2}C + b^{2} – 2ab cosC

Taking out a^{2 }as a common factor, we get;

c^{2} = a^{2}(sin^{2}C + cos^{2}C) + b^{2} – 2ab cosC

Now from the above equation, you know that,

sin^{2}θ + cos^{2}θ = 1

∴ c^{2} = a^{2} + b^{2} – 2ab cosC

Hence, the cosine law is proved.

## Problem and Solution

Let us understand the concept by solving one of the cosines law problems.

**Problem: **A triangle ABC has sides a=10cm, b=7cm and c=5cm. Now, find its angle ‘x’.

**Solution:**

Consider the below triangle as triangle ABC, where,

a=10cm

b=7cm

c=5cm

By using cosines law,

a^{2} = b^{2} + c^{2} – 2bc cos(x)

Or

cos x = (b^{2} + c^{2} – a^{2})/2bc

Substituting the value of the sides of the triangle i.e a,b and c, we get

cos(x) = (7^{2} + 5^{2} – 10^{2})/(2 × 7 × 5)

cos(x)=(49 + 25 -100)/70

cos(x)= -0.37

It is important to solve more problems based on cosines law formula by changing the values of sides a, b & c and cross-check law of cosines calculator given above.

## Frequently Asked Questions – FAQs

### What is law of cosine?

Law of cosines signifies the relation between the lengths of sides of a triangle with respect to the cosine of its angle.

### What does the cosine law states?

As per the cosine law, if ABC is a triangle and α, β and γ are the angles between the sides the triangle respectively, then we have:

a^{2} = b^{2} + c^{2} – 2bc cos α

b^{2} = a^{2} + c^{2} – 2ac cos β

c^{2} = b^{2} + a^{2} – 2ba cos γ

where a,b, and c are the sides of the triangle.

### When should we use the law of cosines?

The cosine law is used to determine the third side of a triangle when we know the lengths of the other two sides and the angle between them.

### Can we use cosine law for all the triangles?

Law of cosine is not just restricted to right triangles, and it can be used for all types of triangles where we need to find any unknown side or unknown angle.

### How to solve the law of cosines?

Cosine law is basically used to find unknown side of a triangle, when the length of the other two sides are given and the angle between the two known sides. So by using the below formula, we can find the length of the third side:

a^{2} = b^{2} + c^{2} -2bc cos α

Where a is the unknown side, b and c are the known sides of the triangle, and α is the angle between b and c.

### What is the formula to find the angle using cosine law?

The formula to find the unknown angles using cosine law is given by:

cos α = [b^{2} + c^{2} – a^{2}]/2bc

cos β = [a^{2} + c^{2} – b^{2}]/2ac

cos γ = [b^{2} + a^{2} – c^{2}]/2ab

## Law of Cosines

In the previous section, we learned how the Law of Sines could be used to solve oblique triangles

in three different situations (1) where a side and two angles (SAA) were known, (2) where two

angles and the included side (ASA) were known, and (3) the ambiguous case where two sides

and an angle opposite one of the sides (SSA) were known. However, how would we solve

oblique triangles where two sides and the included angle (SAS) or where only the three sides

(SSS) are known? In these two cases, there is not enough information to use the Law of Sines so

we must now use a combination of the Law of Cosines and the Law of Sines to solve the oblique

triangle.

**Definition of the Law of Cosines:**

If A, B, and C are the measures of the angles of an oblique triangle, and a, b, and c are the lengths of the sides opposite the corresponding angles, then the square of a side of the triangle is equal to the sum of the squares of the other two sides minus twice the product of the two sides and the cosine of the included angle.

✅ Math Formulas ⭐️⭐️⭐️⭐️⭐

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