## Law of Sines

The law of sines establishes the relationship between the sides and angles of an oblique triangle(non-right triangle). Law of sines and law of cosines in trigonometry are important rules used for “solving a triangle”. According to the sine rule, the ratios of the side lengths of a triangle to the sine of their respective opposite angles are equal. Let us understand the sine law formula and its proof using solved examples in the following sections.

## What is Law of Sines?

The law of sines relates the ratios of side lengths of triangles to their respective opposite angles. This ratio remains equal for all three sides and opposite angles. We can therefore apply the sine rule to find the missing angle or side of any triangle using the requisite known data.

### Law of Sines: Definition

The ratio of the side and the corresponding angle of a triangle is equal to the diameter of the circumcircle of the triangle. The sine law is can therefore be given as,

a/sinA = b/sinB = c/sinC = 2R

- Here a, b, c are the lengths of the sides of the triangle.
- A, B, and C are the angle of the triangle.
- R is the radius of the circumcircle of the triangle.

## Law of Sines Formula

The law of sines formula is used for relating the lengths of the sides of a triangle to the sines of consecutive angles. It is the ratio of the length of the side of the triangle to the sine of the angle thus formed between the other two remaining sides. The law of sines formula is used for any triangle apart from SAS triangle and SSS triangle. It says,

a/sin A = b/sin B = c/sin C

where,

- a, b, and c are the lengths of the triangle
- A, B, and C are the angles of the triangle.

This formula can be represented in three different forms given as,

- a/sinA = b/sinB = c/sinC
- sinA/a = sinB/b = sinC/c
- a/b = sinA/sinB; a/c = sinA/sinC; b/c = sinB/sinC

**Example: Given a = 20 units c = 25 units and Angle C = 42º. Find the angle A of the triangle.**

**Solution:**

For the given data, we can use the following formula of sine law: a/sinA = b/sinB = c/sinC

⇒ 20/sin A = 25/sin 42º

⇒ sin A/20 = sin 42º/25

⇒ sin A = (sin 42º/25) × 20

⇒ sin A = (sin 42º/25) × 20

⇒ sin A = (0.6691/5) × 4

⇒ sin A = 0.5353

⇒ A = sin^{-1}(0.5363)

⇒ A = 32.36º

**Answer: ∠A = 32.36º**

## Proof of Law of Sines Formula

The law of sines is used to compute the remaining sides of a triangle, given two angles and a side. This technique is known as triangulation. It can also be applied when we are given two sides and one of the non-enclosed angles. But, in some such cases, the triangle cannot be uniquely determined by this given data, called the ambiguous case, and we obtain two possible values for the enclosed angle. To prove the sine law, we consider two oblique triangles as shown below.

In the first triangle, we have:

h/b = sinA

⇒ h = b sinA

In the second triangle, we have:

h/a = sinB

⇒ h = a sinB

Also, sin(180º – B) = sinB

Equalizing the h values from the above expressions, we have:

a sinB = b sinA

⇒ a/sinA = b/sinB

Similarly, we can derive a relation for sin A and sin C.

asinC = csinA

⇒ a/sinA = c/sinC

Combining the above two expressions, we have the following sine law.

a/sinA = b/sinB = c/sinC

**Tips and Tricks on Law of Sines**

- The triangulation technique is used to find the sides of a triangle when two angles and one side of a triangle is known. For this the sine law is helpful.
- This sine law of trigonometry should not be confused with the sine law in physics.
- Further deriving from this sine law we can also find the area of an oblique triangle.

Area of a triangle = (1/2) ab sinC = (1/2) bc sinA = (1/2) ca sinB - Also sine law provides a relationship with the radius R of the circumcircle,a/sinA = b/sinB = c/sinC = 2R
- Cosine law: This proves a relationship between the sides and one angle of a triangle,c
^{2}= a^{2}+ b^{2}– 2ab⋅cos C - Tangent law: This has been derived from the sine law and it gives the relationship between the sides and angles of a triangle.

## Applications of Sine Law

The law of sines finds application in finding the missing side or angle of a triangle, given the other requisite data. The sine law can be applied to calculate:

- The length of the side of a triangle using ASA or AAS criteria.
- The unknown angle of a triangle.
- The area of the triangle.

### Ambiguous Case of Law of Sines

While applying the law of sines to solve a triangle, there might be a case when there are two possible solutions, which occurs when two different triangles could be created using the given information. Let us understand this ambiguous case while solving a triangle using Sine law using the following example.

**Example: If the side lengths of △ABC are a = 18 and b = 20 with ∠A opposite to ‘a’ measuring 26º, calculate the measure of ∠B opposite to ‘b’?**

**Solution:**

Using the sine rule, we have sinA/a = sinB/b = sin26º/18 = sin B/20.

⇒ sin B = (9/10) sin26º or B ≈ 29.149º.

However, note that sin x = sin(180º – x). **∵** A + B < 180º and A + (180º – B) < 180º, another possible measure of B is approximately 180º – 29.149º = 150.851º.

**Think out of the box:**

Find the angles of a triangle if the sides are 12 units, 8.5 units, and 7.2 units respectively.

## Examples Using Law of Sines

**Example 1: Two angles and an included side is∠A = 47º and ∠B = 78º and c = 12.6 units. Find the value of a.**

**Solution:**

Given: ∠A = 47º and ∠B = 78º

∠A + ∠B + ∠C = 180º

⇒ 47º + 78º + ∠C = 180º

⇒ 125º + ∠C = 180º

⇒ ∠C = 180º – 125º

⇒ ∠C = 55º

We shall apply the sine law to find the side of the triangle.

a/sin A = c/sin C

⇒ a/sin 47º = 12.6/sin 55º

⇒ a = 5.62

**Answer: a = 11.24 units **

**Example 2: **It is given ∠A = 47º, ∠B = 78º, and the side c = 6.3. Find the length a.

**Solution:**

To find: Length of a

Given:

∠A = 47º, ∠B = 78º, and c = 6.3.

Since, the sum of all the interior angles of the triangle is 180^{∘, }

Therefore,

∠A + ∠B + ∠C=180º

⇒ 47º + 78º + ∠C = 180º

⇒ ∠C = 55º

Using law of sines formula,

a/sinA = b/sinB = c/sinC

⇒ a/sinA = c/sinC

⇒ a/sin47º = 6.3 / sin55º

⇒ a = 6.3 / sin55º × sin47º

⇒ a = 5.6

**Answer:** a = 5.6

**Example 3: For a triangle, it is given a = 10 units c = 12.5 units and angle C = 42º. Find the angle A of the triangle.**

**Solution:**

To find: Angle A

Given:

a = 10, c = 12.5, and angle C = 42º.

Using law of sines formula,

⇒ a/sinA = b/sinB = c/sinC

⇒ 10/sinA = 12.5/sin 42º

⇒ sin A = 0.5353

⇒ ∠A = 32.36º

**Answer:** **∠A = 32.36º**

## FAQs on Sine Law

### What is Meant by Law of Sines?

The Law of sines gives a relationship between the sides and angles of a triangle. The law of sines in Trigonometry can be given as, a/sinA = b/sinB = c/sinC, where, a, b, c are the lengths of the sides of the triangle and A, B, and C are their respective opposite angles of the triangle.

### When Can We Use Sine Law?

Sine law finds application in solving a triangle, which means to find the missing angle or side of a triangle using the requisite given data. We can use the sine law to find,

- Side of a triangle
- The angle of a triangle
- Area of a triangle

### What is the Sine Rule Formula?

The sine rule formula gives the ratio of the sides and angles of a triangle. The sine rule can be explained using the expression, a/sinA = b/sinB = c/sinC. Here a, b, c are the length of the sides of the triangle, and A, B, C are the angles of the triangle.

### What are the Different Ways to Represent Sine Rule Formula?

Sine law can be represented in the following three ways. These three forms are as given below,

- a/sinA = b/sinB = c/sinC
- sinA/a = sinB/b = sinC/c
- a/b = sinA/sinB; a/c = sinA/sinC; b/c = sinB/sinC

### In Which Cases Can We Use the Sine Law?

The sine law can be used for three purposes as mentioned below,

- To find the length of sides of a triangle
- To find the angles of the triangle
- To find the area of the triangle

### Can Sine Law be Used on a Right Triangle?

The sine law can also be used for a right triangle. sine law can be used in oblique(non-right) as well as in a right triangle to establish a relationship between the ratios of sides and their respective opposite angles.

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

The criteria to use the sine law is to have one of the following sets of data known to us,

- A pair of lengths of two sides of a triangle and an angle.
- A pair of angles of a triangle and the length of one side.

### Does Law of Sines Work With 90 Degrees?

Yes, the law of sines and the law of cosines can be applied to both the right triangle and oblique triangle or scalene triangle to solve the given triangle.

### What is the Law of Sines Ambiguous Case?

The law of sines ambiguous case is the case that occurs when there can be two possible solutions while solving a triangle. Given a general triangle, the following given conditions would need to be fulfilled for the ambiguous case,

- Only information given is the angle A and the sides a and c.
- Angle A is acute, i.e., ∠A < 90°.
- Side a is shorter than side c ,i.e., a < c.
- Side a is longer than the altitude h from angle B, where h = c sin A, .i.e., a > h.

### What are the Applications of the Law of Sines?

The law of sines can be applied to find the missing side and angle of a triangle given the other parameters. To apply the sine rule, we need to know either two angles and one side of the triangle (AAS or ASA) or two sides and an angle opposite one of them (SSA).

## Law of Sines Formula Example

**Example**: If angle B = 21^{0}, angle C= 46^{0} and the side AB = 9 cm in a triangle is given. Find the other sides of triangle.

**Solution:**

Given: two angles and a side

Let’s use the Sine rule to solve this.

As the sum of angles in a triangle is 180^{0}

Accordingly, angle A = 113^{0}

As AB = c = 9 cm.

Use the Sine Rule:

### Sample Problems

**Problem 1: Find remaining lengths of the triangle XYZ when ∠X = 30° and ∠Y = 45° and x = 5 cm.**

**Solution:**

Given data, ∠X = 30°, ∠Y = 45° and x = 5 cm

We know that sum of the three angles of a triangle is 180°

So, ∠X + ∠Y + ∠Z = 1

30° + 45° + ∠Z = 180° ⇒ 75° + ∠Z = 180°

∠Z = 105°

Now from the law of Sines, x/ sin X = y/ sin Y = z/ sin Z

x/sin X = y/ sin Y ⇒ 5/ sin 30° = y/ sin 45°

⇒ x/(1/2) = y/(1/√2) ⇒ 10 =√2y ⇒ y = 7.07 cm

Similarly, x/ sin X = z/sin Z ⇒ 5/sin 30° = z/sin 105° [sin 105° = (√6 + √2)/4 = 0.965]

⇒ 5/(1/2) = z/(0.965) ⇒** z = 9.65 cm**

**Problem 2: Find ∠P and ∠Q and the length of the third side when ∠R = 36° and p = 2.5 cm and r = 7 cm?**

**Solution:**

Given, ∠R = 36° and p = 2.5 cm and r = 7 cm

From the Sine Rule Formula we have,

p/sin P = q/sin Q = r/sin R

⇒ 2.5/sin P = q/ sin Q = 7/sin 36°

⇒ 2.5/sin P = 7/sin 36° [sin 36° = 0.5878]

⇒ sin P = 0.20992 ⇒ P = sin^{-1 }(0.20992)

**⇒ ∠P = 12.12°**

⇒ We have, ∠P + ∠Q + ∠R = 180°

⇒12.12° + ∠Q + 36° = 180° ⇒ **∠Q = 131.88°**

⇒ q/sin 131.88° = 7/sin 36°

⇒ q/0.7445 = 7/0.5878 ⇒ **q = 8.866 cm **(approximately)

Hence ∠P = 12.12°, ∠Q = 131.88° and q = 8.866 cm

**Problem 3: Find the ratio of the sides of the triangle ABC when ∠A = 15°, ∠B = 45°, and ∠C = 120°?**

**Solution:**

Given: ∠A =15°, ∠B = 45° and ∠C = 120°

From the Sine Rule formula, **a/ sin A = b/ sin B = c/ sin C ⇒ a : b : c = sin A : sin B : sin C**

⇒ a : b : c = sin 15° : sin 45° : sin 120°

sin 15° = (√3 – 1)/2√2 = 0.2588 (approximate value)

sin 45° = 1/√2 = 0.7071 (approximate value)

sin 120° = √3/2 = 0.866 (approximate value)

Hence the ratio of the three sides of the triangle ABC is **a : b : c = 0.2588 : 0.7071 : 0.866**

**Problem 4: Find the area of the triangle ABC when BC = 10 cm, AB = 12 cm and ∠B= 30°?**

**Solution:**

Given, BC = a = 10 cm, AB = c = 12 cm and ∠B= 30°

We know that, **Area of the triangle = ½ (base) (height) = ½ (a) (h)** ⇢ (1)

From the figure, sin B = height/c

h = c sin B ⇢ (2)

Now substitute equation (2) in (1),

**Area of the triangle ABC = ½ (a)(c) sin B**= ½ (10) (12) sin 30° [sin 30° = ½]

⇒ Area = ½ (120) ½ = 30 cm^{2}

Hence the area of the triangle ABC is** 30 cm ^{2}.**

**Problem 5: Find ∠ACB if a = 3 cm, c = 1 cm, and ∠BAC = 60°.**

**Solution:**

Given, ∠BAC = 60°, a = 3 cm and c = 1 cm

From the Sine Rule Formula, we have **sin A/a = sin C/c**

⇒ sin 60°/3 = sin C/1 [sin 60° = √3/2]

⇒ (√3/2)/3 = sin C/3 ⇒ sin C = 1/(2√3) ⇒ sin C = 0.2887 (approximately)

⇒ ∠C = sin^{-1}(0.2887) ⇒ ** ∠C = 16.77°**

Hence** ∠ACB = 16.77°**

**Problem 6: Find the length of the side YZ, if the area of the triangle XYZ is 24 cm ^{2}, ∠Y = 45°, and z = 4 cm.**

**Solution:**

Given, Area of the triangle XYZ = 24 cm

^{2}, ∠y = 45° and z = 4 cm

From the Sine rule law, we have

Area of triangle XYZ = ½ (x)(z) sin Y⇒ 24 = ½ (x)(4) sin 45° [sin 45° = 1/√2]

⇒ 24 = (x) × (2) × (1/√2)

⇒ 12√2 = x ⇒

x = 16.968 cmHence the

length of the side YZ = x = 16.968 cm

**Problem 7: Find q when ∠P= 108°, ∠Q = 43°and p = 15 cm?**

**Solution:**

Given, ∠P = 108° , ∠Q = 43° and p = 15 cm

From the Sine Rule Formula we have,** p/sin P = q/sin Q**

⇒ 15/sin 108° = q/sin 43°

⇒ 15/(0.9510) = q/(0.6820) [sin 108° = 0.9510 & sin 43° = 0.6820]

⇒ q = 10.757 cm (approximately)

Hence **q =10.57 cm**

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