Sine and Cosine Rules (HSC SSCE Mathematics Advanced): Revision Notes

Sine and Cosine Rules

Learning objectives

Sine rule and area of triangle formula

Understanding the sine rule

The sine rule is a formula that relates the sides of a triangle to the sines of its angles. It is particularly useful for solving problems involving non-right-angled triangles.

The sine rule states:

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

where:

• $a$, $b$, and $c$ are the lengths of the sides
• $A$, $B$, and $C$ are the angles opposite to sides $a$, $b$, and $c$ respectively

Proof of the sine rule

To understand where the sine rule comes from, we can construct a proof using right-angled triangles.

Consider a triangle with sides $a$, $b$, and $c$ opposite angles $A$, $B$, and $C$ respectively. We draw an altitude (perpendicular line) from vertex $C$ to the opposite side $AB$, meeting at point $D$. This altitude, labelled $h$, divides the triangle into two right-angled triangles.

Using the definition of sine in right-angled triangles:

In $\triangle ACD$:

$\sin A = \frac{h}{b}$

Therefore:

$h = b \sin A$

In $\triangle BCD$:

$\sin B = \frac{h}{a}$

Therefore:

$h = a \sin B$

Since both expressions equal $h$, we can equate them:

$b \sin A = a \sin B$

In a non-degenerate triangle, $\sin A \neq 0$ and $\sin B \neq 0$, so we can divide both sides by $\sin A \sin B$:

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

A similar argument can be applied by constructing an altitude from another vertex (for example, from $A$ to $BC$), which gives:

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

Combining both results leads to the complete sine rule:

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

Area of triangle formula

The area of a triangle can be calculated using two sides and the included angle between them.

For a triangle $ABC$ with sides $a$ and $b$ about the angle $C$, the area formula is:

$A = \frac{1}{2}ab \sin C$

where:

• $A$ is the area of the triangle
• $a$ and $b$ are the sides adjacent to angle $C$
• $C$ is the included angle

Proof of the area formula

Consider triangle $ABC$ with base $a$ (side $BC$) and height $h$ from vertex $A$.

The standard area formula for any triangle is:

$A = \frac{1}{2} \times \text{base} \times \text{height}$

In this triangle, the base is $a$ and the height is $h = b \sin C$.

Substituting these values:

$A = \frac{1}{2} \times a \times b \sin C$

$A = \frac{1}{2}ab \sin C$

Similarly, depending on which base and height you use, the area can also be expressed as:

$A = \frac{1}{2}bc \sin A$

or

$A = \frac{1}{2}ac \sin B$

Summary of formulas

Formula	Expression
Sine rule	$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
Area of triangle formula	$A = \frac{1}{2}ab \sin C$

To apply the sine rule:

• Know one angle-side pair and either another angle (to find a side) or another side (to find an angle)
• Verify the smallest angle faces the shortest side

To compute area:

• Use two sides and the included angle

Ambiguous case of the sine rule

What is the ambiguous case?

The ambiguous case in trigonometry refers to using the sine rule to calculate the size of an angle in a triangle where there are two possibilities for the angle, one obtuse and one acute. This leads to two possible triangles.

The ambiguous case occurs because, for an acute angle $\theta$, we have:

$\sin(180° - \theta) = \sin \theta$

This means that both an acute angle and its supplementary obtuse angle have the same sine value.

When does the ambiguous case occur?

The ambiguous case arises in SSA triangles (two sides and a non-included angle) when using the sine rule. In this configuration, $\sin x = \sin(180° - x)$ may yield two possible angles for a solution.

For triangle $ABC$ with known $a$, $b$, and $\angle A$ (acute), the ambiguous case occurs if:

Finding both solutions

When the ambiguous case produces two valid triangles, we calculate both angles using:

$B_1 = \sin^{-1}\left(\frac{b \sin A}{a}\right)$

$B_2 = 180° - B_1$

where:

• $B_1, B_2$ are the possible angles opposite side $b$ in triangle $ABC$
• $a, b$ are the known side lengths in SSA configuration
• $A$ is the known angle in SSA configuration

Geometric construction of the ambiguous case

Fix $\angle A$ and side $b$ at vertex $A$. Draw side $a$ from point $B$. Point $B$ can lie at two positions ($B_1, B_2$), forming angles $x$ and $180° - x$, if $b \sin A \leq a < b$.

This creates two different triangles:

• Possibility 1: Triangle with vertices $A$, $C_1$, $B_1$
• Possibility 2: Triangle with vertices $A$, $C_2$, $B_2$ where $B_2 = 180° - B_1$

Cosine rule

Understanding the cosine rule

The cosine rule is a formula that relates the lengths of the sides of a triangle to the cosine of one of its angles.

$c^2 = a^2 + b^2 - 2ab \cos C$

where:

• $c$ is the side opposite to angle $C$
• $a, b$ are the sides adjacent to angle $C$

Finding sides and angles with the cosine rule

To find a side:

$c^2 = a^2 + b^2 - 2ab \cos C$

where:

• $c$ is the side opposite to angle $C$
• $a, b$ are the sides adjacent to $C$

To find an angle:

$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$

where:

• $c$ is the side opposite to angle $C$
• $a, b$ are the other two sides

Proof of the cosine rule

Consider a triangle with sides $a$, $b$, and $c$ opposite angles $A$, $B$, and $C$ respectively. To establish the cosine rule, we construct an altitude from vertex $B$ to the opposite side $AC$, intersecting at point $D$. This altitude $h$ divides the triangle into right-angled triangles, $\triangle ABD$ and $\triangle BCD$.

Let $CD = x$, so that $AD = b - x$.

Using Pythagoras' theorem:

In $\triangle BCD$:

$h^2 + x^2 = a^2$

In $\triangle ABD$:

$h^2 + (b - x)^2 = c^2$

Expanding the second equation:

$h^2 + b^2 - 2bx + x^2 = c^2$

Substituting $h^2 = a^2 - x^2$ from the first equation:

$a^2 - x^2 + b^2 - 2bx + x^2 = c^2$

Simplifying:

$a^2 + b^2 - 2bx = c^2$

In $\triangle BCD$, which has a right angle at point $D$, we know that:

$\cos C = \frac{x}{a}$

since $x$ is the adjacent side and $a$ is the hypotenuse.

Therefore:

$x = a \cos C$

Substituting into our equation:

$a^2 + b^2 - 2ab \cos C = c^2$

Thus, the cosine rule is given by:

$c^2 = a^2 + b^2 - 2ab \cos C$

Remember!