**Law of sines **defines the ratio of sides of a triangle and their respective sine angles are equivalent to each other. This law is used to find an unknown angle or unknown sides. This is one of the two trigonometric function laws apart from the law of cosines.

Law of sines can be used for all types of triangles such as an acute, obtuse and right triangle. Let us learn here how to prove this law and its uses with the help of examples.

## Law of Sines Definition

In general, the law of sines is defined as the ratio of side length to the sine of the opposite angle. It holds for all the three sides of a triangle respective of their sides and angles.

**\(\frac{a}{Sin A}=\frac{b}{Sin B}=\frac{c}{Sin C}\)**

So, we use the Sine rule to find unknown lengths or angles of the triangle. It is also called as **Sine Rule**, **Sine Law** or **Sine Formula.Â **

**Also, read:**

Law Of Cosines | Law Of Tangents |

Trigonometric Functions | Trigonometric Identities |

Trigonometry | Inverse Trigonometric Functions |

## Formula

The formulas used with respect to law of sine are given below.

a / Sin A= b/ Sin B= c / Sin C |

a: b: c = Sin A: Sin B: Sin C |

a / b = Sin A / Sin B |

b / c = Sin B / Sin C |

It denotes that if we divide side a by the Sine of âˆ A, it is equal to the division of side b by the Sine ofâˆ B and also equal to the division of side c by Sine of âˆ C (Or)Â The sides of a triangle are to one another in the same ratio as the sines of their opposite angles.

Here, Sin A is a number and a is the length.

## Applications

The trigonometry ratios such as sine, cosine and tangent are primary functions which are used to find the unknown angles or sides of a triangle. The applications of sine law are given below:

- It can be used to compute the other sides of a triangle when two angles and one side is given.
- When two sides and one non-included angle is given.

The sine law is usually used to find the unknown angle or unknown angle.

As per the law, we know, if a, b and c are the lengths of three sides of a triangle and âˆ A, âˆ B and âˆ C are the angles between the sides, then;

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

Now if suppose, we know the value of one of the side, and the value of two angles, such as:

a = 7cm, âˆ A = 60Â°, âˆ B = 45Â°

find b?

By the sine law, we know;

a/sin A = b/sin B

Now putting the values, we get;

7/sin 60Â° = b/sin 45Â°

7/(âˆš3/2) = b/(1/âˆš2)

14/âˆš3 =Â âˆš2 b

## Law of Sines Proof

We need a right-angled triangle to prove the above as the trigonometric functions are mostly defined in terms of this type of triangle only.

**Given:** â–³ABC.

**Construction:** Draw a perpendicular, CD âŠ¥ AB. Then CD = h is the height of the triangle. “h”Â separates the â–³ ABC in two right-angled triangles, â–³CDA and â–³CDB.

**To Show:** a / b = Sin A / Sin B

**Proof:** In the Â â–³CDA,

Sin A= h/b

And in Â â–³CDB,

Sin B = h/a

Therefore, Sin A / Sin B = (h / b) / (h / a)=Â a/b

And we proved it.

Similarly, we can prove, Sin B/ Sin C= b / c and so on for any pair of angles and their opposite sides.

## Law of Sines-Ambiguous Case

In an ambiguous case, if in a triangle two sides and the angle opposite to them are known, then there could be three possibilities:

- That there is no such triangle
- There are two different triangle
- Or there is exactly one such triangle

## Problems and Solutions

**Question:Â Solve â–³PQR in which âˆ P = 63.5Â° and âˆ Q = 51.2Â° Â and r = 6.3 cm.**

Solution:

Letâ€™s Calculate the third angle-

âˆ RÂ = 180Â° – 63.5Â°- 51.2 Â°= 65.3Â°

Now, Letâ€™s calculate the sides:

6.3/Sin 65.3 = p/Sin 63.5 (BC= p)

p = (6.3 Ã— Sin 63.5)/Sin 65.3

p= 6.21 cm approximately.

Similarly, 6.3/Sin 65.3 = q/Sin 51.2 (q = AB)

q = (6.3 Ã— Sin51.2) / Sin 65.3

q= 5.40 cm

âˆ R= 65.3Â°

p=6.21 cm

q=5.40 cm

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