3.2 Triangles - The Sine Rule And The Cosine Rule

1. Basic Trigonometric Notions

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

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

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

The Pythagorean Identity:

Utilizing Pythagoras' theorem ($a^2 = b^2 + c^2$), the fundamental mathematical identity is derived directly:

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

EXAMPLE 1

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

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

To calculate the numerical angle $B$, inverse trigonometric functions (e.g., $\sin^{-1}$) are required. $B = \sin^{-1}(0.8) \approx \mathbf{53.1^\circ}$. Because the angles inside a triangle total $180^\circ$, the remaining acute angle calculates strictly to $90^\circ - 53.1^\circ = \mathbf{36.9^\circ}$.

Values for Basic Angles

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

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

2. The Sine Rule and The Cosine Rule

For any generic triangle with sides $a, b, c$ and opposing angles $A, B, C$, two universal geometric rules apply:

The Sine Rule:
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
The Cosine Rule:
$a^2 = b^2 + c^2 - 2bc\cos A$
Alternative permutations shift relative to the target angle: $b^2 = a^2 + c^2 - 2ac\cos B$.

EXAMPLE 2 (Verifying Right Triangles)

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

Sine Rule validates: $\frac{a}{\sin 90^\circ} = \frac{b}{\sin B} \Rightarrow a = \frac{b}{\sin B} \Rightarrow \sin B = \frac{b}{a}$.
Cosine Rule validates: $a^2 = b^2 + c^2 - 2bc\cos 90^\circ \Rightarrow a^2 = b^2 + c^2$ (Pythagoras theorem recovered identically because $\cos 90^\circ = 0$).

3. The Solution of a Triangle

A triangle consists of 6 core elements (3 sides, 3 angles). Given any 3 elements (excluding 3 angles), the remaining elements are solvable structurally using logical rule selection:

  • Use Cosine Rule: When Three Sides (SSS) or Two Sides and an Included Angle (SAS) are known.
  • Use Sine Rule: When an angle and its directly opposite side are known.

EXAMPLE 3 (Given Three Sides)

Evaluate angles for a triangle with sides $a=4, b=3, c=2$.

Cosine Rule for A: $4^2 = 2^2 + 3^2 - 12\cos A \Rightarrow 16 = 13 - 12\cos A \Rightarrow \cos A = -0.25 \Rightarrow \mathbf{A \approx 104.5^\circ}$.
Cosine Rule for B: $3^2 = 2^2 + 4^2 - 16\cos B \Rightarrow 9 = 20 - 16\cos B \Rightarrow \cos B = 0.6875 \Rightarrow \mathbf{B \approx 46.6^\circ}$.
Angle C evaluates automatically: $180^\circ - 104.5^\circ - 46.6^\circ = \mathbf{28.9^\circ}$.

EXAMPLE 4 (Given SAS)

Evaluate side $BC$ for a triangle where $AB=3, AC=2$, and included angle $A=104.5^\circ$.

Cosine Rule applies: $BC^2 = 2^2 + 3^2 - 12\cos(104.5^\circ) \approx 16$.
Therefore $BC = \mathbf{4}$.

EXAMPLE 6 & 7 (The Ambiguous Case)

Providing two sides and a non-included angle triggers the Ambiguous Case via the Sine Rule, yielding zero, one, or potentially two valid geometric triangles.

Single Solution (Ex 6): Given $a=2, b=3$, and $B=46.6^\circ$.
$\frac{3}{\sin 46.6^\circ} = \frac{2}{\sin A} \Rightarrow \sin A \approx 0.484$. Solving evaluates $A \approx 28.9^\circ$. The supplementary angle $151.1^\circ$ is geometrically rejected because $151.1^\circ + 46.6^\circ > 180^\circ$.
Two Solutions (Ex 7): Given $c=5, b=4$, and $B=30^\circ$.
$\frac{4}{\sin 30^\circ} = \frac{5}{\sin C} \Rightarrow \sin C = 0.625$.
Two distinct valid geometries emerge:
Case 1: $C = \mathbf{38.7^\circ}$, forcing $A = 111.3^\circ$ and producing $BC = 7.45$.
Case 2: $C = 180^\circ - 38.7^\circ = \mathbf{141.3^\circ}$, forcing $A = 8.7^\circ$ and producing $BC = 1.21$.
No Solution (Ex 8): Given $c=5, b=1$, and $B=30^\circ$.
$\frac{1}{\sin 30^\circ} = \frac{5}{\sin C} \Rightarrow \sin C = 2.5$. This is mathematically impossible since sine bounds limit outputs to $[-1, 1]$. No triangle exists.

4. The Area of a Triangle

Using SAS criteria, the surface area evaluates strictly via:

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

EXAMPLE 9

Given the triangle from Example 3 ($a=4, b=3, c=2, A=104.5^\circ$), calculating area relies on any corner pair:

$\text{Area} = \frac{1}{2}(2)(3)\sin(104.5^\circ) \approx \mathbf{2.90}$