Trigonometry and Triangle Geometry for the ESAT

Triangle Notation and Basic Area

To solve trigonometric problems effectively, we must use a consistent system for labelling triangles. Generally, we label the vertices with capital letters $A, B, C$ and the sides opposite those vertices with the corresponding lower case letters $a, b, c$. While diagrams often label corners anticlockwise, this is not a universal rule, so you must always verify the relative positions of sides and angles.

The most fundamental formula for the area of a triangle is derived from the area of a rectangle. As shown below, a triangle with a horizontal base and a vertical height is exactly half the area of the rectangle enclosing it.

$area = \frac{1}{2} \times base \times vertical\ height = \frac{1}{2} bh$

This formula remains valid even if the top corner is not directly above the base. In such cases, you must use the perpendicular vertical height above the line of the horizontal base, rather than the slanted height of the side of the triangle.

The Trigonometric Formula for Area

We can refine our area calculation using trigonometry. Consider a triangle where we know two sides and the angle between them.

In this diagram, the base is $a$. The vertical height can be found using the right-angled triangle formed by the slant height $b$ and the angle $C$. The height is $b \sin C$. Substituting this into our basic area formula gives:

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

This formula is powerful because it allows us to calculate the area of any triangle provided we have two sides and the included angle. By rotating the triangle, we can see that the area is also equal to $\frac{1}{2}bc \sin A$ and $\frac{1}{2}ca \sin B$.

The Sine Rule

Because the area of a triangle is constant regardless of which sides and angles we use to calculate it, we can equate the different forms of the area formula:

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

By multiplying this entire equation by 2 and then dividing by $abc$, we derive the sine rule:

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

Alternatively, we can use the reciprocal form, which is often easier when solving for a side length:

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

When to use the Sine Rule

1. Given two angles and one side: You can find any other side. Note that once you have two angles, you can find the third by subtracting from 180 degrees.
2. Given two sides and one angle (not the angle between the sides): You can find another angle.

The Ambiguous Case of the Sine Rule

When we use the sine rule to find an angle (the side, side, angle case), we may encounter an ambiguous result. This happens because for any value of $\sin \theta = k$, there are two possible angles between 0 and 180 degrees: $\theta$ and $180 - \theta$.

Worked Example: Ambiguity

In triangle $ABC$, let angle $A = 30^{\circ}$, side $b = 6$, and side $a = 4$. Find angle $B$.

Using the sine rule:

$\frac{\sin B}{6} = \frac{\sin 30^{\circ}}{4}$

$\sin B = \frac{6 \sin 30^{\circ}}{4} = \frac{6 \times 0.5}{4} = \frac{3}{4}$

This gives two possible values for $B$: $48.6^{\circ}$ or $180 - 48.6 = 131.4^{\circ}$. Since both $30 + 48.6$ and $30 + 131.4$ are less than 180, both triangles are theoretically possible. This ambiguity typically occurs when the side opposite the given angle is shorter than the other given side.

Sine Rule and Circle Geometry

Every triangle can be inscribed in a circle, known as the circumcircle. The radius of this circle, $R$, is related to the sine rule.

By using circle theorems, specifically that the angle at the centre is twice the angle at the circumference, it can be proven that:

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

This shows that the ratio of any side to the sine of its opposite angle is always equal to the diameter of the circumcircle.

The Cosine Rule

The cosine rule generalizes Pythagoras' theorem for triangles that do not have a right angle. In a right-angled triangle, $a^2 = b^2 + c^2$.

If we distort the triangle, we must add a correction term to keep the equation balanced.

Using Pythagoras on the sub-triangles in the diagram above:

1. $(b + x)^2 + y^2 = a^2$
2. $x^2 + y^2 = c^2$

Substituting equation 2 into equation 1 yields $a^2 = b^2 + 2bx + c^2$. Since $x = -c \cos A$, we arrive at the standard cosine rule:

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

This can be written for any side by permuting the letters:

$b^2 = c^2 + a^2 - 2ca \cos B$

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

If you need to find an angle when three sides are known, rearrange the formula:

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

When to use the Cosine Rule

1. Given three sides: You can find any angle.
2. Given two sides and the included angle: You can find the third side.

Three Dimensional Problems

In the ESAT, you may encounter these rules in 3D contexts, such as finding the angle between a line and a plane or distance between vertices of a pyramid. The strategy is to identify 2D triangles within the 3D shape and apply the sine or cosine rules to those specific planes.