3.2 Sine Rule and Cosine Rule

1. Basic Trigonometric Notions

For a right-angled triangle, the sine, cosine, and tangent of an angle $\theta$ are defined by the ratios of its sides:

$\sin\theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}}$ $\cos\theta = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}$ $\tan\theta = \dfrac{\text{Opposite}}{\text{Adjacent}}$

This establishes the identity: $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$.

The Pythagorean Identity:

Using Pythagoras' theorem ($a^2 = b^2 + c^2$), the identity is derived:

$$\sin^2\theta + \cos^2\theta = \left(\dfrac{b}{a}\right)^2 + \left(\dfrac{c}{a}\right)^2 = \dfrac{b^2 + c^2}{a^2} = \dfrac{a^2}{a^2} = \mathbf{1}$$

EXAMPLE 1

Consider a right-angled triangle with sides 3, 4, and hypotenuse 5. For angle $B$ opposite the side of length 4:

$\sin B = \dfrac{4}{5} = 0.8$
$\cos B = \dfrac{3}{5} = 0.6$
$\tan B = \dfrac{4}{3} \approx 1.333$

To find angle $B$, use inverse trigonometric functions (e.g., $\sin^{-1}$):

$$B = \sin^{-1}(0.8) \approx \mathbf{53.1^\circ}$$

Since the angles inside a triangle total $180^\circ$, the remaining acute angle is $90^\circ - 53.1^\circ = \mathbf{36.9^\circ}$.

Values for Basic Angles

$\theta$	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$
$\sin\theta$	$0$	$\dfrac{1}{2}$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{\sqrt{3}}{2}$	$1$
$\cos\theta$	$1$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{1}{2}$	$0$
$\tan\theta$	$0$	$\dfrac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$	Undefined

Note: The sine values follow the pattern $\dfrac{\sqrt{0}}{2}, \dfrac{\sqrt{1}}{2}, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{3}}{2}, \dfrac{\sqrt{4}}{2}$. Supplementary angles have equal sines ($150^\circ$ and $30^\circ$) but opposite cosines.

2. The Sine Rule and The Cosine Rule

For a triangle with sides $a, b, c$ and opposite angles $A, B, C$, two rules apply:

The Sine Rule:

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

The Cosine Rule:

$$a^2 = b^2 + c^2 - 2bc\cos A$$

Alternative forms: $b^2 = a^2 + c^2 - 2ac\cos B$.

EXAMPLE 2 (Verifying Right Triangles)

Applying the rules to a right-angled triangle ($A = 90^\circ$) verifies their definitions:

The Sine Rule yields:

$$\begin{aligned} \dfrac{a}{\sin 90^\circ} &= \dfrac{b}{\sin B} \\ \dfrac{a}{1} &= \dfrac{b}{\sin B} \\ \sin B &= \mathbf{\dfrac{b}{a}} \quad \text{(Opposite / Hypotenuse)} \end{aligned}$$

The Cosine Rule yields:

$$\begin{aligned} a^2 &= b^2 + c^2 - 2bc\cos 90^\circ \\ a^2 &= b^2 + c^2 - 2bc(0) \\ \mathbf{a^2} &= \mathbf{b^2 + c^2} \quad \text{(Pythagorean theorem derived)} \end{aligned}$$

3. The Solution of a Triangle

A triangle has 3 sides and 3 angles. Given any 3 elements (except three angles), the remaining elements can be found:

Use the Cosine Rule: When three sides (SSS) or two sides and an included angle (SAS) are known.
Use the Sine Rule: When an angle and its opposite side are known.

EXAMPLE 3 (Given Three Sides: SSS)

Find the angles of a triangle with sides $a=4, b=3, c=2$.

Cosine Rule for A:

$$\begin{aligned} 4^2 &= 2^2 + 3^2 - 2(2)(3)\cos A \\ 16 &= 4 + 9 - 12\cos A \\ 16 &= 13 - 12\cos A \\ 12\cos A &= -3 \\ \cos A &= -0.25 \implies \mathbf{A \approx 104.5^\circ} \end{aligned}$$

Cosine Rule for B:

$$\begin{aligned} 3^2 &= 2^2 + 4^2 - 2(2)(4)\cos B \\ 9 &= 4 + 16 - 16\cos B \\ 9 &= 20 - 16\cos B \\ 16\cos B &= 11 \\ \cos B &= 0.6875 \implies \mathbf{B \approx 46.6^\circ} \end{aligned}$$

Angle C is calculated directly:
$$180^\circ - 104.5^\circ - 46.6^\circ = \mathbf{28.9^\circ}$$

EXAMPLE 4 (Given SAS)

Find side $BC$ ($a$) for a triangle where $AB=3$ ($c$), $AC=2$ ($b$), and included angle $A=104.5^\circ$.

Use the Cosine Rule to find the opposite side:

$$\begin{aligned} a^2 &= b^2 + c^2 - 2bc\cos A \\ BC^2 &= 2^2 + 3^2 - 2(2)(3)\cos(104.5^\circ) \\ BC^2 &= 4 + 9 - 12(-0.25) \\ BC^2 &= 13 + 3 \\ BC^2 &= 16 \end{aligned}$$

Therefore $BC = \mathbf{4}$.

EXAMPLE 5, 6 & 7 (The Ambiguous Case)

Two sides and a non-included angle can produce zero, one, or two valid triangles (the ambiguous case).

Single Solution (Ex 5): Given $a=2, b=3$, and $B=46.6^\circ$.

$$\begin{aligned} \dfrac{3}{\sin 46.6^\circ} &= \dfrac{2}{\sin A} \\ \sin A &= \dfrac{2\sin 46.6^\circ}{3} \approx 0.484 \end{aligned}$$

Thus, $A \approx 28.9^\circ$. The supplementary angle $151.1^\circ$ is rejected because $151.1^\circ + 46.6^\circ > 180^\circ$.

Two Solutions (Ex 6): Given $c=5, b=4$, and $B=30^\circ$.

$$\begin{aligned} \dfrac{4}{\sin 30^\circ} &= \dfrac{5}{\sin C} \\ \sin C &= \dfrac{5\sin 30^\circ}{4} = \dfrac{5(0.5)}{4} = 0.625 \end{aligned}$$

Two distinct cases exist because both the acute and obtuse angles form valid triangles:

Case 1: $C = \sin^{-1}(0.625) = \mathbf{38.7^\circ}$, so $A = 180^\circ - 30^\circ - 38.7^\circ = 111.3^\circ$.
Using Sine Rule for $a$: $BC = \dfrac{4\sin 111.3^\circ}{\sin 30^\circ} = \mathbf{7.45}$.
Case 2: $C = 180^\circ - 38.7^\circ = \mathbf{141.3^\circ}$, so $A = 180^\circ - 30^\circ - 141.3^\circ = 8.7^\circ$.
Using Sine Rule for $a$: $BC = \dfrac{4\sin 8.7^\circ}{\sin 30^\circ} = \mathbf{1.21}$.

No Solution (Ex 7): Given $c=5, b=1$, and $B=30^\circ$.

$$\begin{aligned} \dfrac{1}{\sin 30^\circ} &= \dfrac{5}{\sin C} \\ \sin C &= \dfrac{5\sin 30^\circ}{1} = 5(0.5) = 2.5 \end{aligned}$$

This is impossible because sine values cannot exceed 1. No triangle exists because side $b$ is too short.

4. The Area of a Triangle

Given two sides and the included angle (SAS), the area is calculated using the sine formula:

$$\text{Area} = \dfrac{1}{2}bc\sin A = \dfrac{1}{2}ab\sin C = \dfrac{1}{2}ac\sin B$$

EXAMPLE 8

For the triangle from Example 3 ($a=4, b=3, c=2, A=104.5^\circ$), the area is:

$$\begin{aligned} \text{Area} &= \dfrac{1}{2}bc\sin A = \dfrac{1}{2}(2)(3)\sin(104.5^\circ) \\ &= 3 \times \sin(104.5^\circ) \approx \mathbf{2.90} \end{aligned}$$