Law of Sines Triangle

The law of sines is the relationship between angles and sides of all types of triangles such as acute, obtuse and right-angle triangles. It states the ratio of the length of sides of a triangle to sine of an angle opposite that side is similar for all the sides and angles in a given triangle.

To use the law of sines, you need to know either two sides of an angle of a triangle ( AAS or ASA)  or two sides and an angle opposite one of them ( SSA). In this article, you will learn the law of sines, the law of sine definition, the law of sine formula, the law of sine problems, the law of sine proof, the law of sine applications and when to use law of sines.

Law of Sine Definition

The law of sine is defined as the ratio of the length of sides of a triangle to the sine of the opposite angle of a triangle. The law of sine is also known as Sine rule, Sine law, or Sine formula.

Law of sine is used to solve traingles.

\[\frac{a}{\text{Sin a}} = \frac{b}{\text{Sin b}} = \frac{c}{\text{Sin c}}\]

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a, b, and c are sides of the above triangle whereas A, B, and C are angles of above triangle. 

( Sie a is opposite to angle A, side b is opposite to angle B, side c is opposite the angle C.)

(When we will divide side a by the sine of angle A, it will be equal to the side b by the sine of angle b and it will also be equivalent to side c by the sine of angle C).

Law of Sines Formula

The law of sine formula to find the unknown sides and angles of the triangle is given below

• \[\frac{p}{\text{Sin P}} = \frac{q}{\text{Sin Q}} = \frac{r}{\text{Sin R}}\]  

• p :q : r = Sin P : Sin Q: Sin R

• \[\frac{p}{q} = \frac{\text{Sin P}}{\text{Sin Q}}\]

• \[\frac{q}{r} = \frac{\text{Sin Q}}{\text{Sin R}}\]

The above law of sine formula represents that if we will divide side a by the sine of angle A, it will be equal to the side b by the sine of angle b and it will also be equivalent to side c by the sine of angle C).

Here, Sin A represents a number whereas a is the length of a triangle.

Law of Sine Proof

For the law of sine proof, we need to have right-angle triangles as trigonometry functions are mostly defined in terms of right-angle triangles only.

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Given: △ ABC

Construction: Draw a perpendicular line i.e. CD⊥ AB,  then CD = height of the triangle. “H” divides the ABC into two right-angled triangles.

To proof : \[\frac{a}{b} = \frac{\text{Sin a}}{\text{Sin b}}\]

Proof: In △ CDA

Sin A  = \[\frac{h}{b}\]

And in △ CDB

Sin B  = \[\frac{h}{a}\]

Hence, \[\frac{\text{Sin A}}{\text{Sin B}} = \frac{(\frac{h}{b})}{(\frac{h}{a})} = \frac{a}{b}\]

Hence, proved

Similarly, we can prove \[\frac{Sin B}{Sin C} = \frac{b}{c}\] and other corresponding pairs of angles and their opposite sides.

Law of Sine - Ambiguous Case

If in a triangle any two sides and their opposite angles are known, then there are three possibilities in the law of sine- ambiguous case. M these three possibilities are:

1. That there is no such triangle

2. There are two distinct triangle

3. There is exactly one such triangle

When to Use the Law of Sines?

The law of sine is used to find the angles of an ordinary triangle. In two sides and the enclosed angles are given, it can be simultaneously used to find the third side and other two angles.

If two sides and angles opposite one of them are known then the other can be calculated.

The third angle is then calculated by the fact that the sum of the interior angles of a triangle is equal to 180 degrees.

We use the following law of sine formula to calculate the unknown sides and angles of a triangle.

\[\frac{a}{\text{Sin a}} = \frac{b}{\text{Sin b}} = \frac{c}{\text{Sin c}}\]   

Law of sine problem

Here, you can see the law of sine problem with the solution

1. Calculate △PQR in which ∠ P = 116°, p = 8.3 cm and q - 5.4 cm.

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Solution: \[\frac{Sin 116^{\circ}}{8.3}\] = \[\frac{SinQ}{5.4}\]

Sin Q = \[5.4\frac{Sin 116^{\circ}}{8.3}\] = 0.5848

Q = = Sin-¹ 0.5848

35.8° or 144.2°

Q cannot be an obtuse angle as the sum of the interior angle of a triangle will exceed 

180°. The only value for Q will be considered is 35.8°.

∠Q = 35.8° ,  ∠ R  = 180° - 116° - 35.8° = 28.2°

\[\frac{8.3}{Sin 116^{\circ}}\] = \[\frac{r}{Sin 28.2^{\circ}}\]

r = \[8.3\frac{Sin 28.2^{\circ}}{Sin 116^{\circ}}\]  = 4.36

Hence, in the above triangle ∠Q = 35.8°, ∠ R = 28.2° and side r = 4.36 cm

Quiz Time

1. If, c= 6cm, b = 9 cm and a = 14 cm. Find A

1. 133

2. 137

3. 188

4. 121

2. In triangle ABC, if ∠A = 98° and ∠B = 12°. If side a is 84 units long approximately then find the length of side b.

1. 18 units

2. 17 units

3. 12 units

4. 86 units

Solved Examples

Here, you can see some of  the law of sine example problem

1. If  ∠B = 21°, ∠c = 46 and side AB = 9 cm in a triangle then solve the triangle.

Solution:

Given: Two angles and a side

No, let’s use law of sine to solve the triangle

As we know sum of the interior sides of a triangle is 180°

Accordingly, ∠A = 113°

As AB = c= 9cm

Using the law of sine rule:

\[\frac{a}{Sin 113^{\circ}} = \frac{b}{Sin 21^{\circ}} =  \frac{9}{Sin 46^{\circ}}\]

\[\frac{b}{Sin 21^{\circ}} = \frac{9}{Sin 46^{\circ}}\]

b = Sin 21° × \[\frac{9}{Sin 46^{\circ}}\]

= 4.484 cm

A = sin 113° ×  \[\frac{9}{Sin 46^{\circ}}\]

= 11.517 cm.

2. In PQR, Sin p = 1/3 and Sin q = ¼, Find the ratio of side p/side q

Solution:

Using the law of sine, we get

\[\frac{p}{\text{Sin P}} = \frac{q}{\text{Sin Q}} = \frac{p}{\frac{1}{3}} = \frac{q}{\frac{1}{4}}\]

\[\frac{1}{3p}\] = \[\frac{1}{4q}\]

\[\frac{4}{1}\times \frac{1}{3q} = p\]

\[\frac{4}{3} = \frac{p}{q}\]

Hence, the ratio of the side p to side q is 4 to 3.