Law of Sine - Definition, Proof, Formula, Applications and Example

Whether it is navigation, architecture, medical imaging, or computer graphics, principles of trigonometry are applicable in various fields of science and technology. The presence of triangles is very evident around us in the shape of a field, a mountain side, a flag, a pizza slice, etc. The solution of triangles involves the application of properties, rules, and theorems to analyze them more deeply. Some of these rules are the Sine rule, Cosine rule, Tangent rule, Projection rule, etc. These rules can be applied to analyze triangles involved in different cases. In real life, we use the Sine rule to measure the navigation and angle of tilt.

In this article, we will cover the concept of Sine rule. This category falls under the broader category of Trigonometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of seventeen questions have been asked on this concept, including two in 2022.

Triangle

A triangle is a polygon with 3 sides. A triangle is more special as compared to other polygons as it is the polygon having the least number of sides. A triangle has six main elements, three sides, and three angles. There are different rules and theorems for triangles that relate their sides and angles.

A few standard symbols to represent elements of the triangle:
In $\triangle A B C$, the angles are denoted by capital letters $A, B$, and $C$, and the length of the sides opposite to these angles are denoted by small letters $a, b$, and $c$ respectively.

The following symbols in relation to $\triangle A B C$ are universally adopted.
Angles: $\angle B A C=A, \angle A B C=B, \angle B C A=C$
Sides: $A B=c, A C=b$, and $B C=a$
Semi-perimeter of the $\triangle \mathrm{ABC}$, is $s=\frac{a+b+c}{2}$ and it is denoted by s . So, the perimeter of $\triangle A B C$ is $2 s=a+b+c$.

The area of a triangle is denoted by $S$ or $\Delta$.
For any $\triangle A B C$,
- $A+B+C=180^{\circ}$
- $a+b>c, b+c>a$ and $c+a>b$
- $a>0, b>0, c>0$

Law of sine

The law of sine states that "In a triangle, side "a" divided by the sine of angle $A$ is equal to side " $b$ " divided by the sine of angle $B$ is equal to side " $c$ " divided by the sine of angle $C$."

Sine Rule

The law of sine rule connects the angle of a triangle with the length of its sides. It is used to find the unknown length or angle of a triangle.

The law of sine rule states that:

In any triangle, The ratio of the sine of one of the angles to the length of its opposite side will be equal to the other two ratios of the sine of the angle measured to the opposite side.

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=2 \mathrm{R}$

Where $R$ is the circumradius of a triangle.

Proof:
Using the right triangle relationships, we know that

$
\sin A=\frac{h}{b} \text { and } \sin B=\frac{h}{a}
$
Solving both equations for h gives two different expressions for h

$
\mathrm{h}=\mathrm{b} \sin \mathrm{A} \text { and } \mathrm{h}=\mathrm{a} \sin \mathrm{B}
$
We then set the expressions equal to each other.

$
b \sin A=a \sin B
$

$\left(\frac{1}{a b}\right)(b \sin A)=(a \sin B)\left(\frac{1}{a b}\right) \quad$ Multiply both sides by $\frac{1}{a b}$

$
\frac{\sin A}{a}=\frac{\sin B}{b}
$
Similarly, we can compare the other ratios.

$
\frac{\sin A}{a}=\frac{\sin C}{c} \text { and } \frac{\sin B}{b}=\frac{\sin C}{c}
$
Collectively,

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

Applications of Sine Rule

Trigonometry ratios such as sine, cosine, and tangent are used to find the side of a triangle if the angle of the triangle is given.

The applications of the sine rule are:

1) We can use the sine rule to find the length of the side of a triangle if two angles and one side length are given.

2) With the help of the sine rule we can find the angle of a triangle if two sides length and one angle is given.

Problems Based on the Sine rule

Example 1: Let in a right-angled triangle, the smallest angle be $\theta$. If a triangle formed by taking the reciprocal of its sides is also a right-angled triangle, then $\sin \theta$ is equal to :
[JEE MAINS}

Solution

Here

$
b<a<\sqrt{a^2+b^2}
$
Their reciprocals:

$
\frac{1}{\sqrt{a^2+b^2}}<\frac{1}{a}<\frac{1}{b}
$

These also form a right-angled triangle, so

$\begin{aligned} & \therefore \frac{1}{a^2+b^2}+\frac{1}{a^2}=\frac{1}{b^2} \\ & \Rightarrow \frac{a^2+a^2+b^2}{a^2\left(a^2+b^2\right)}=\frac{1}{b^2} \\ & \Rightarrow 2 a^2 b^2+b^4=a^4+a^2 b^2 \\ & \Rightarrow b^4+a^2 b^2-a^4=0\end{aligned}$
$
\begin{aligned}
& \Rightarrow\left(\frac{b}{a}\right)^4+\left(\frac{b}{a}\right)^2-1=0 \quad\left(\text { dividing by } a^4\right) \\
& \Rightarrow\left(\frac{b}{a}\right)^2=\frac{-1 \pm \sqrt{1+4}}{2}=\frac{\sqrt{5}-1}{2} \text { (negative rejected) }
\end{aligned}
$
$
\text { From the first figure } \frac{b}{a}=\tan \Theta
$
$
\begin{aligned}
& \Rightarrow \tan ^2 \theta=\frac{\sqrt{5}-1}{2} \\
& \Rightarrow \sec ^2 \theta=\frac{\sqrt{5}+1}{2} \Rightarrow \cos ^2 \theta=\frac{2}{\sqrt{5}+1} \\
& \Rightarrow \sin ^2 \theta=\frac{\sqrt{5}-1}{\sqrt{5}+1} \Rightarrow \sin \theta=\frac{\sqrt{5}-1}{2}
\end{aligned}
$Hence the correct answer is $\frac{\sqrt{5-1}}{2}$

Example 2: Let $\frac{\sin A}{\sin B}=\frac{\sin (A-C)}{\sin (C-B)}$, where $A, B, C$ are the angles of a triangle $A B C$. If the lengths of the sides opposite these angles are $a, b, c$ respectively, then :
[JEE MAINS 2021]

Solution

$\begin{aligned} & \frac{\sin A}{\sin B}=\frac{\sin (A-C)}{\sin (C-B)} \\ & A=\pi-B-C, B=\pi-A-C \\ & \Rightarrow \frac{\sin (B+C)}{\sin (A+C)}=\frac{\sin (A-C)}{\sin (C-B)} \\ & \Rightarrow \sin (C+B) \sin (C-B)=\sin (A+C) \sin (A-C) \\ & \Rightarrow \sin ^2 C-\sin ^2 B=\sin ^2 A-\sin ^2 C \\ & \Rightarrow \sin ^2 A, \sin ^2 C, \sin ^2 B \text { are in } A P \\ & \Rightarrow a^2, c^2, b^2 \text { are in } A P\end{aligned}$
$
\Rightarrow a^2, c^2, b^2$ are in A P

Hence, the answer is $b^2, c^2, a^2$ are an $A . P$.

Example 3 : If $A B C$ is a triangle and $\$$ mathrm $\{A B\}=2, B C=\ \operatorname{sqrt}\{3\} \$$ then what is the maximum possible value of $\$ A \$$ ?

Solution
In $\triangle A B C$, the angles are denoted by capital letters $A, B$, and $C$, and the length of the sides opposite to these angles are denoted by small letters a, b, and c respectively.

$
\begin{aligned}
& \angle B A C=A \\
& \angle A B C=B \\
& \angle B C A=C
\end{aligned}
$
Sides of the $\triangle A B C$

$
A B=c, A C=b \text {, and } B C=a
$
Collectively, these relationships are called the Law of Sines

$
\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}
$
$
\begin{aligned}
& \frac{A B}{\sin C}=\frac{B C}{\sin A} \\
& \frac{2}{\sin C}=\frac{\sqrt{3}}{\sin A} \\
& \sin A=\frac{\sqrt{3}}{2} \sin C
\end{aligned}
$

maximum possible value of $\mathrm{A}=\frac{\pi}{3}$

Hence, the answer is $\frac{\pi}{3}$

Example 4:If ABC is the triangle and $A-B=90^{\circ}$ and $\mathrm{B}=3 \mathrm{~A}$ then Find the value of $B$ .
Solution
In $\triangle A B C$, the angles are denoted by capital letters $A, B$, and $C$, and the length of the sides opposite to these angles are denoted by small letters $\mathrm{a}, \mathrm{b}$ and $c$ respectively.

$\begin{aligned} \angle B A C & =A \\ \angle A B C & =B \\ \angle B C A & =C\end{aligned}$

Sides of the $\triangle A B C$ $A B=c, A C=b$, and $B C=a$ $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

$\begin{gathered}\frac{a}{\sin A}=\frac{b}{\sin B} \\ \frac{b}{3 \sin (90+B)}=\frac{b}{\sin B}\end{gathered}$

Hence, the answer is 3.

Example 5: In triangle PQR, $p^2, q^2$ and $r^2$ are in $A . P$. Then $\cot P, \cot Q, \cot R$ are in
Solution

$\begin{aligned} & p^2, q^2, r^2 \text { are in } A . P \\ & r^2-q^2=q^2-p^2 \\ & \sin ^2 R-\sin ^2 Q=\sin ^2 Q-\sin ^2 P \\ & \sin (R+Q) \sin (R-Q)=\sin (Q+P) \sin (Q-P) \\ & \sin (\pi-P)(\sin R \cos Q-\cos R \sin Q)=\sin (\pi-R)(\sin Q \cos P-\cos Q \sin P) \\ & \sin (P)(\sin R \cos Q-\cos R \sin Q)=\sin (R)(\sin Q \cos P-\cos Q \sin P) \\ & \text { divide both } \operatorname{side} \text { by } \sin P \sin Q \sin R \\ & \cot Q-\cot R=\cot P-\cot Q \\ & \cot P, \cot Q, \cot R \text { are in A.P. }\end{aligned}$

Hence the answer is $\cot P, \cot Q$, and $\cot R$ are in $A P$.

Summary

The sine rule is a powerful tool in trigonometry that provides a straightforward method for solving triangles and understanding their geometric properties. Its versatility and simplicity make it a fundamental concept in both mathematics and applications involving geometry and trigonometry