Heron's formula (derivation using trig): Foundations

This page assumes you have seen nothing. We build every symbol, every picture, and every idea the parent note leans on, in an order where each new thing only uses things already built.

1. What is a triangle, and what are $a$ , $b$ , $c$ ?

A triangle is a flat shape made of three straight sticks (sides) joined end to end, enclosing some space.

We give the three sides names — the letters $a$ , $b$ , $c$ — and we give the three corners (called vertices) the capital letters $A$ , $B$ , $C$ .

Figure — Heron's formula (derivation using trig)

Why the topic needs this: Heron's derivation constantly mixes a side ( $c$ ) with the corner facing it (angle $C$ ). The Law of Cosines only works because of this pairing. Look at the figure: side $c$ (coral) and angle $C$ (lavender wedge) are on opposite ends of the triangle — remember that pairing and the whole derivation stays organised.

2. What is an angle, and what is $C$ ?

An angle measures how much you must turn to swing from one side to the other side meeting at a corner. A tiny wedge = a nearly-closed pair of sticks; a wide wedge = sticks spread far apart.

$C$ is the angle at corner $C$ — it lives between the two sides that touch there, namely sides $a$ and $b$ . Because $a$ and $b$ both end at corner $C$ , angle $C$ is called the angle included between $a$ and $b$ .

Why the topic needs this: The first line of the whole derivation is $Area = \frac{1}{2} ab sin C$ . That $C$ is precisely the wedge between the two named sides $a$ and $b$ . If you grab the wrong angle, the formula lies. The figure shows the wedge sitting exactly in the gap between the mint and butter sides.

3. Height and area — what "space inside" means

The area of a shape is how much flat space it covers. For a triangle we normally compute it as

$Area = \frac{1}{2} \times base \times height .$

base = any one side you choose to sit the triangle on.
height = the straight-down distance from the opposite corner to that base (a perpendicular drop, meeting the base at a right angle).

Why the topic needs this: The parent note's whole point is to get the area without measuring the height — because in real life you often can't reach the top of the triangle with a ruler. Understanding what "height" normally means is what makes Heron's height-free result feel magical.

4. The sine of an angle — $sin C$

Here is the tool that lets us find area from two sides and their angle, skipping the height.

Picture side $a$ lying flat as the base. Side $b$ leans off corner $C$ at angle $C$ . The height of the triangle is how high the far tip of $b$ rises above the base. That rise equals $b sin C$ — the length $b$ scaled by "how steeply it points up."

Why $sin$ and not something else? We want the vertical rise off the base. $sin$ is precisely the "how-much-goes-up" ratio for a given angle — that is its whole job. This is the Area of a triangle = ½ab sin C launching point of the derivation.

All cases of $sin C$ for a real triangle (angle $C$ strictly between $0°$ and $180°$ ):

If $C = 0°$ : the sticks lie flat on top of each other, $sin 0° = 0$ , area $= 0$ . Degenerate — no triangle.
If $C = 90°$ : sticks are perpendicular, $sin 90° = 1$ , maximum rise. Right triangle.
If $C \to 180°$ : sticks stretch into a straight line, $sin 180° = 0$ , area $\to 0$ again.

So for any genuine triangle, $0 < C < 180°$ and $sin C > 0$ — the area is always positive. Good.

5. The cosine of an angle — $cos C$

$cos C$ is the partner of $sin C$ . Where $sin C$ measures the "up" part of the leaning stick, $cos C$ measures the "along the base" part.

Why the topic needs both: The derivation's clever move is to trade the hard-to-pin-down $sin C$ for $cos C$ , because $cos C$ can be written using only side lengths (next section). The identity $sin^{2} C = 1 - cos^{2} C$ is the bridge between them. See Law of Cosines.

6. Squares of lengths and $c^{2}$

$c^{2}$ just means $c \times c$ — the length $c$ multiplied by itself. Geometrically it is the area of a square whose side is $c$ .

Why the topic needs this: Squaring is exactly why the derivation squares the area (to remove the awkward $sin$ ) — it keeps everything in the "square of a length" world where Pythagoras and the Law of Cosines live.

7. Square root —

The square root $x$ answers the question: "what positive number, multiplied by itself, gives $x$ ?" It undoes squaring.

$36 = 6$ because $6 \times 6 = 36$ .
$x$ only makes sense (as a real number) when $x \geq 0$ : you cannot un-square into a negative.

Why the topic needs this: Heron's formula ends in $s (s - a) (s - b) (s - c)$ . The derivation first finds $Area^{2}$ (easier, because squaring cleared the $sin$ ), then takes the square root at the very end to recover the actual area. The " $x \geq 0$ needed" rule is exactly what forces the triangle inequality — see §10.

8. The semi-perimeter $s$

The perimeter is the total distance around the triangle: $a + b + c$ . The semi-perimeter is half of that:

The strange-looking pieces $s - a$ , $s - b$ , $s - c$ each have a clean meaning: $s - a = \frac{b + c - a}{2}, s - b = \frac{a + c - b}{2}, s - c = \frac{a + b - c}{2} .$ Each one is "half of (the two other sides minus this side)."

Why the topic needs $s$ and not the full perimeter: the algebra of the derivation naturally produces factors like $(a + b - c)$ , which are twice the $(s - c)$ pieces. Packaging them as $s$ makes the final formula short and symmetric. See Semi-perimeter and Incircle (r = Area/s).

9. Difference of squares — the key algebra move

Why the topic needs it: In the derivation, $1 - cos^{2} C$ is a difference of squares ( $1^{2} - cos^{2} C$ ), and after substituting sides you get numerators like $(a + b)^{2} - c^{2}$ — again differences of squares. Splitting them produces the neat factors $(a + b + c)$ , $(a + b - c)$ , etc., which are exactly the $s$ -pieces. See Difference of Squares Factoring.

10. The triangle inequality — when sides actually make a triangle

Three lengths only form a real triangle if no side is longer than the other two combined:

Rewriting with $s$ : these three conditions are exactly $s - a > 0$ , $s - b > 0$ , $s - c > 0$ .

Why the topic needs it: Recall from §7 that needs a non-negative inside. Heron's inside is $s (s - a) (s - b) (s - c)$ . The triangle inequality is precisely what guarantees all three $(s - \cdot)$ factors are positive, so the root is real. Fail it (e.g. $1, 1, 5$ ) and $s - c < 0$ : Heron correctly refuses. See Triangle Inequality.

Prerequisite map

Equipment checklist

Test yourself — cover the right side and answer before revealing.

In a triangle, which side is opposite corner

C

Side

c

— small letters face matching capitals.

Which two sides is the angle

C

included between?

Sides

a

and

b

(both meet at corner

C

What does

sin C

physically measure on the triangle?

The "up" ratio — how much the leaning side rises off the base; height

= b sin C

Why is

sin C > 0

for every real triangle?

Because

0 < C < 180°

, and sine is positive across that whole range.

What is

cos C

measuring instead?

The "along the base" part of the leaning side; negative when

C

is obtuse.

State the Pythagorean identity linking them.

sin^{2} C + cos^{2} C = 1

, so

sin^{2} C = 1 - cos^{2} C

What does

c^{2}

mean geometrically?

The area of a square with side length

c

Write the Law of Cosines.

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

What does

x

do and when is it real?

Undoes squaring; real only when

x \geq 0

Define the semi-perimeter

s

s = \frac{a + b + c}{2}

, half the perimeter.

Factor

X^{2} - Y^{2}

(X - Y) (X + Y)

State the triangle inequality in

s

-form.

s - a > 0

s - b > 0

s - c > 0

(equivalently

a + b > c

etc.).

Connections

Area of a triangle = ½ab sin C — the area tool built from $sin C$ .
Law of Cosines — turns $cos C$ into pure side lengths.
Difference of Squares Factoring — the algebra that cracks $1 - cos^{2} C$ .
Triangle Inequality — makes the square root real.
Semi-perimeter and Incircle (r = Area/s) — where $s$ reappears.
Heronian Triangles — triangles where everything comes out an integer.
Heron's formula (derivation using trig) — the parent this page prepares you for.