Law of Sine is a basic law of trigonometry that defines the relation between the sides and the angles of the triangle. It is used to express the relation between the sides and the angles of the triangle. It is also known as Sine Law or Sine Rule or Sine Formula. The formula for the Law of Sines is expressed as:
a/sin A = b/sin B = c/sin C
Here, a, b, and c are the sides of the triangle, and A, B, and C are the angles opposite these sides.
Law of Sines is versatile and can be applied in various fields such as navigation, surveying, and engineering. It helps calculate the heights of inaccessible objects, like buildings or trees, by measuring angles and distances from different points.
Law of Sine Definition
Law of sine is the ratio of the side length to the sine of the opposite angles. For a triangle with sides, a, b, and c with respective angles, ∠A, ∠B, and ∠C the sine law states that,
a/sin A = b/sin B = c/sin C
The law of sine formula is used to find the relation between the lengths of a triangle's sides to the sines of consecutive angles. It is the ratio of the length of the triangle's side to the sine of the angle generated by the other two remaining sides.
It is given by :
a/sin A = b/sin B = c/sin C
Where,
- a, b, and c are Lengths of Triangle
- A, B, and C are Angles of Triangle
Proof of the Law of Sines
Law of sine is proved using the triangles added below,
To derive Sine Law, consider two oblique triangles illustrated above,
sin A = h/b
⇒ h = b sin A . . . (1)
In Right Triangle,
sin B = h/a
⇒ h = a sin B . . .(2)
From (1) and (2), we get
a sin B = b sin A
⇒ a/sinA = b/sinB
Similarly,
a/sinA = c/sinC
By combining the two formulas above,
a/sin A = b/sin B = c/sin C
The is Law of Sine formula.
Sine Formula is the other name of the Law of Sine, or Sine Rule. The sine formulas are various formulas that are related to the side and angle of the triangle. The formulas listed below are used to solve various trigonometric problems:
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 |
|---|
Applications of Sine Law
Some of the important applications of sine law are :
- It is used to find the length and angle of a triangle when sides or angles are given.
- It is used to solve various trigonometric problems.
- It is used in finding the height and distances of various real-life cases, etc.
Law of Sines and Cosines
Law of Sine and Cosine relate the sine and cosine of an angle of the triangle with the other angles or sides of the triangle. Let us suppose we are given a triangle ABC with sides a, b, and c and angles A, B, and C respectively then,
Law of Sine is given by :
a/SinA = b/SinB = c/SinC
Law of Cosine is given by,
a2 = b2 + c2 − 2bc cos A
or, b2 = a2 + c2 − 2ac cos B
or, c2 = a2 + b2 − 2ab cos C
Related :
Examples on Law of Sines
Example 1: It is given for a triangle ABC, a = 20 units, c = 25 units, and ∠C = 30°. Find ∠A of the triangle.
Solution:
Given,
- a = 20 units
- c = 25 units
- ∠C = 30°
Using Sine Formula
a/sin A = c/sin C
20/sin A = 25/sin 30
sin A = 0.40
A = 23.5°
Example 2: It is given for a triangle ABC, b = 15 units, c = 20 units, and ∠C = 60°. Find ∠B of the triangle.
Solution:
Given,
- b = 15 units
- c = 20 units
- ∠C = 60°
Using Sine Formula
b/sin B = c/sin C
15/sin B = 20/sin 60
sin B = 0.649448
B = 40.5°
Example 3: It is given for a triangle ABC, b = 30 units, c = 40 units, and ∠C = 30º. Find ∠B of a triangle.
Solution:
Given,
- b = 30 units
- c = 10 units
- ∠C = 30°
Using Sine Formula
b/sin B = c/sin C
30/sin B = 40/sin 30
sin B = 0.374607
B = 22°
Example 4: It is given for a triangle ABC, a = 15 units, b = 20 units, and ∠C = 45°. Find ∠A of the triangle.
Solution:
Given,
a = 15 units
b = 20 units
∠C = 45°
Using Sine Formula
a/sin A = b/sin B
15/sin A = 20/sin 45
sin A = 0.75
A = 48.6°
Example 5: It is given for a triangle ABC, a = 10 units, b = 14 units, and ∠A = 30°. Find ∠B of the triangle.
Solution:
Given,
a = 10 units
b = 14 units
∠A = 30°
Using Sine Formula
a/sin A = b/sin B
10/sin 30 = 14/sin B
sin B = 0.7
B = 44.4°
Example 6: It is given for a triangle ABC, a = 25 units, b = 30 units, and ∠C = 50°. Find ∠A of the triangle.
Solution:
Given,
a = 25 units
b = 30 units
∠C = 50°
Using Sine Formula
a/sin A = b/sin B
25/sin A = 30/sin 50
sin A = 0.6
A = 36.9°
Example 7: It is given for a triangle ABC, b = 18 units, c = 24 units, and ∠A = 60°. Find ∠B of the triangle.
Solution:
Given,
b = 18 units
c = 24 units
∠A = 60°
Using Sine Formula
b/sin B = c/sin C
18/sin B = 24/sin 60
sin B = 0.674
B = 42.1°
Example 8: It is given for a triangle ABC, a = 12 units, b = 16 units, and ∠C = 90°. Find ∠A of the triangle.
Solution:
Given,
a = 12 units
b = 16 units
∠C = 90°
Using Sine Formula
a/sin A = b/sin B
12/sin A = 16/sin 90
sin A = 0.75
A = 48.6°
Example 9: It is given for a triangle ABC, b = 25 units, c = 30 units, and ∠C = 40°. Find ∠B of the triangle.
Solution:
Given,
b = 25 units
c = 30 units
∠C = 40°
Using Sine Formula
b/sin B = c/sin C
25/sin B = 30/sin 40
sin B = 0.616
B = 38.0°
Example 10: It is given for a triangle ABC, a = 9 units, c = 11 units, and ∠C = 80°. Find ∠A of the triangle.
Solution:
Given,
a = 9 units
c = 11 units
∠C = 80°
Using Sine Formula
a/sin A = c/sin C
9/sin A = 11/sin 80
sin A = 0.801
A = 53.2°
Law of Sines Practice Problems
1. If in a triangle with sides, a = 8, b = 7, and angle ∠A = 120° are given. Find the corresponding value of ∠B.
2. In a triangle with sides, a = 12, b = 9, and angle ∠A = 90° are given. Find the corresponding value of ∠B.
3. For a triangle of sides a = 6, b = 4, and angle ∠A = 60° are given. Find the corresponding value of ∠B.
4. In a triangle of sides, a = 18, b = 12, and angle ∠A = 30° are given. Find the corresponding value of ∠B.
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