## Introduction to the Law of Sines

Law of sines may be used in the technique of triangulation to find out the unknown sides when two angles and a side are provided. It is also applicable when two sides and one unenclosed side angle are given. We may use the form to find out unknown angles in a scalene triangle. The Law of sine and cosine are the trigonometric equations that are used to calculate unknown lengths and angles for a scalene triangle.

### Define Law of Sines

For triangles, like we have the law of cosines, we have the law of sines. The Law of sines is a trigonometric equation where the lengths of the sides are associated with the sines of the angles related. The law of sines is described as the side length of the triangle divided by the sine of the angle opposite to the side. The formula for the sine rule of the triangle is:

\[\frac{a}{sin A}\] = \[\frac{b}{sin B}\] = \[\frac{c}{sin C}\] (where a, b, c are sided lengths of the triangle and A, B, C are opposite angles to the respective sides)

Therefore, side length a divided by the sine of angle A is equal to side length b divided by the sine of angle B is equal to side length c divided by the sine of angle C.

It may also be written as \[\frac{a}{sin A}\] = \[\frac{b}{sin B}\] = \[\frac{c}{sin C}\] = d (where d stands for the diameter of the triangle’s circumcircle).

\[\frac{sin A}{a}\] = \[\frac{sin B}{b}\] = \[\frac{sin C}{c}\] is also an acceptable form of the law of sines.

## Law of Sines Proof

### Sine Rule Proof

To derive the law of sines, let us take the area of a triangle whose sides are a, b, c and the angles opposite to the respective sides are A, B, and C.

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Area = \[\frac{1}{2}\]b(c sin A) = \[\frac{1}{2}\] a(b sin C) = \[\frac{1}{2}\]c(a sin B)

Multiplying by \[\frac{2}{abc}\],

\[\frac{2T}{abc}\] = \[\frac{sin A}{a}\] = \[\frac{sin B}{b}\] = \[\frac{sin C}{c}\]

Hence, \[\frac{sin A}{a}\] = \[\frac{sin B}{b}\] = \[\frac{sin C}{c}\]

Alternative method,

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Let us take a triangle ΔABC where O is the circumcentre, D lies on BC such that OD is perpendicular to BC.

ΔBDO ≅ ΔCDO

CD = BD = \[\frac{a}{2}\]

∠BOD = ∠COD

∠BOC = 2∠BAC,

Let ∠BAC = θ,

∴ ∠BOD = ∠COD = θ

In ΔBOD,

sinθ = \[\frac{\frac{a}{2}}{R}\]

\[\frac{a}{sin \theta }\] = 2R

Similar scenarios occur for \[\frac{b}{sin \theta }\] = 2R and \[\frac{c}{sin \theta}\] = 2R.

### The Formula For Sine Rule

As discussed earlier, the set of formulae for sine rule are:

\[\frac{a}{sin A}\] = \[\frac{b}{sin B}\] = \[\frac{c}{sin C}\]

\[\frac{sin A}{a}\] = \[\frac{sin B}{b}\] = \[\frac{sin C}{c}\]

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

\[\frac{a}{b}\] = \[\frac{sin A}{sin B}\]

\[\frac{b}{c}\] = \[\frac{sin B}{sin C}\]

\[\frac{c}{a}\] = \[\frac{sin C}{sin A}\]

The sine rule states that if we divide the side length of any side of a triangle with the sine of its opposite angle, it is equal to the ratio of another side and the sine of its opposite angle. This implies that the sides of the triangle are in the same proportion to that of the sine of their respective opposite angles.

## Solved Examples

1. Calculate the Value of AB for a Triangle ΔABC Such that Angle C is 105⁰ and Angle B is 35⁰ and side AC=7 cm.

Ans. We observe that we have been given two angles and one side to calculate another side.

Given,

∠B = 35⁰

∠C = 105⁰

AC = b = 7

AB = c = ?

Now we may apply the law of sines,

\[\frac{a}{sin A}\] = \[\frac{b}{sin B}\] = \[\frac{c}{sin C}\]

\[\frac{b}{sin B}\] = \[\frac{c}{sin C}\]

c = \[\frac{b sin C}{sin B}\]

c = \[\frac{7 \times sin 105^{0}}{sin 35^{0}}\]

c = \[\frac{7 \times 0.966}{0.574}\]

c = 11.78 ≈ 11.8 cm

Hence, the side AB is measured as 11.8cm.

2. For a Triangle, Given AB=5.5 and AC=4.7. Angle C is Given to be 63⁰. Find Angle B of the Triangle.

Ans. We observe that we have been given two sides and one angle to calculate another angle.

Given,

AB = c = 5.5

AC = b = 4.7

∠C = 63⁰

∠B = ?

Applying the law of sines,

\[\frac{a}{sin A}\] = \[\frac{b}{sin B}\] = \[\frac{c}{sin C}\]

\[\frac{b}{sin B}\] = \[\frac{c}{sin C}\]

sin B = \[\frac{b sin C}{c}\]

sin B = \[\frac{4.7 \times sin 63^{0}}{5.5}\]

sin B = \[\frac{4.7 \times 0.89}{5.5}\]

sin B = 0.76

sin\[^{-1}\](sin B) = sin\[^{-1}\] 0.76

B = 49.46⁰

Hence angle B is 49.46⁰.

1. What are Some of the Applications of the Law of Sines?

Ans. The law of sines and the law of cosines hold a very important place in trigonometry. The basic trigonometric ratios or functions like sine, cosine are extremely useful in finding out unknown values or parameters in a triangle. The law of sines is extremely helpful to determine the unknown parameter of a side of a triangle when two angles and one of those opposite sides are provided. It is also applicable when two of the sides and one unenclosed angle is provided.

The Law of sines is one of the most commonly used methods along with the law of cosines to find the unknown angles and sides of a scalene triangle. The sine rule can be further extended and generalized to bigger dimensions including a constant curvature. Law of sines in further combinations can be used in proving congruence of triangles using ASA and AAS criteria.

2. Use a Right-angled Triangle to Prove the Law of Sines.

**Law of Sines Right Triangle:**

Let us take a ΔABC. The trigonometric ratios sine, cosine, etc. are easily applicable in a right-angled triangle. So, we need to divide the triangle into two right-angled triangles. Taking point D on AB such that CD is perpendicular to AB. Now we form two right-angled triangles ΔCDA and ΔCDB.

BC = a

AC = b

AB = c

We want to show that a/sin A = b/sin B.

In ΔCDA,

sin A = CD/b

In ΔCDB,

sin B = CD/a

Hence, a/sin A = b/sin B.

Similarly, we may prove that b/sin B = c/sin C.

Therefore we prove the law of sines as a/sin A = b/sin B = c/sin C.