Extended Triangle Laws

🔺 Extended Triangle Laws: The Secret Tools of Triangle Masters

Imagine you’re a detective with a magnifying glass, but instead of solving crimes, you’re solving triangles! These seven powerful formulas are like seven magic keys that unlock any triangle’s secrets.

🎯 The Big Picture: Why Do We Need More Laws?

You already know the Sine Rule and Cosine Rule. But what if you only know certain pieces of information? What if the usual formulas don’t fit your puzzle?

That’s where these Extended Triangle Laws come in. Think of them as specialized tools in your math toolbox — each designed for a specific job.

1️⃣ Projection Formula: The Shadow Rule

What Is It?

Imagine the sun is directly above one side of a triangle. The projection is like the shadow that the other sides cast onto that base.

The Formula

For a triangle with sides a, b, c opposite to angles A, B, C:

a = b·cos(C) + c·cos(B)
b = c·cos(A) + a·cos(C)
c = a·cos(B) + b·cos(A)

Simple Analogy

Think of a ladder leaning against a wall. The ladder’s “shadow” on the ground is the projection. Each side of a triangle “projects” onto another side!

Example

In a triangle where:

• b = 5, c = 7
• B = 60°, C = 40°

Find side a:

a = 5 × cos(40°) + 7 × cos(60°)
a = 5 × 0.766 + 7 × 0.5
a = 3.83 + 3.5
a = 7.33

When to Use It

✅ When you know two sides and the angles opposite to them ✅ To verify your answers from other methods

2️⃣ Napier’s Analogy: The Tangent Shortcut

What Is It?

John Napier (the guy who invented logarithms!) found a clever way to connect the difference and sum of two sides to the tangent of half-angles.

The Formula

tan[(A-B)/2]   (a-b)
───────────── = ─────
tan[(A+B)/2]   (a+b)

Also written as:

(a-b)/(a+b) = tan[(A-B)/2] / tan[(A+B)/2]

Simple Analogy

Imagine comparing two siblings’ heights. Napier’s formula tells you: “The ratio of their height difference to their total height relates to half the difference of their ‘angle personalities’!”

Example

Given: a = 8, b = 5, and C = 50°

Since A + B + C = 180°, we have A + B = 130°

(a-b)/(a+b) = tan[(A-B)/2] / tan(65°)

(8-5)/(8+5) = tan[(A-B)/2] / 2.145

3/13 = tan[(A-B)/2] / 2.145

tan[(A-B)/2] = 0.495

(A-B)/2 = 26.3°

So A - B = 52.6°

Combined with A + B = 130°:

• A = 91.3°
• B = 38.7°

When to Use It

✅ When you know two sides and the included angle (SAS) ✅ To find individual angles quickly

3️⃣ Mollweide’s Formula: The Ultimate Checker

What Is It?

This formula connects all three sides and all three angles in one beautiful equation. It’s perfect for checking your work!

The Formulas

(a-b)/c = sin[(A-B)/2] / cos(C/2)

(a+b)/c = cos[(A-B)/2] / sin(C/2)

Simple Analogy

Think of Mollweide’s formula as the “balance checker” at a bank. After you count all your coins (sides and angles), this formula makes sure everything adds up correctly!

Example

Verify a triangle with a = 7, b = 5, c = 6, A = 78.5°, B = 44.4°, C = 57.1°

Check using the first formula:

(a-b)/c = (7-5)/6 = 0.333

sin[(A-B)/2] / cos(C/2)
= sin(17.05°) / cos(28.55°)
= 0.293 / 0.879
= 0.333 ✓

When to Use It

✅ To verify all your calculated values ✅ When you have complete triangle data

4️⃣ Law of Tangents: The Balanced Equation

What Is It?

Similar to Napier’s Analogy, but written in a slightly different form that some find easier to remember.

The Formula

(a-b)/(a+b) = tan[(A-B)/2] / tan[(A+B)/2]

This can also be written as:

(a-b)·tan[(A+B)/2] = (a+b)·tan[(A-B)/2]

Simple Analogy

It’s like a seesaw! The difference of sides times one angle-tangent equals the sum of sides times another angle-tangent. Perfect balance!

Example

Given: a = 10, b = 6, C = 60°

Step 1: Find A + B = 180° - 60° = 120°

Step 2: Apply the formula

(10-6)/(10+6) = tan[(A-B)/2] / tan(60°)

4/16 = tan[(A-B)/2] / 1.732

tan[(A-B)/2] = 0.433

(A-B)/2 = 23.4°

Step 3: Solve

• A - B = 46.8°
• A + B = 120°
• A = 83.4°, B = 36.6°

When to Use It

✅ Perfect for SAS problems ✅ When you want to find angles from sides

5️⃣ m-n Theorem: The Cevian Master

What Is It?

When a line (called a cevian) is drawn from a vertex to the opposite side, dividing it in ratio m:n, this theorem tells you the relationship between angles.

The Setup

graph TD
    A((A)) --- B((B))
    A --- C((C))
    B --- C
    A --- D((D))
    style D fill:#ff6b6b

Point D divides BC in ratio m:n (BD:DC = m:n)

The Formula

(m+n)·cot(θ) = m·cot(B) - n·cot(C)
              = n·cot(β) - m·cot(α)

Where:

• θ = angle ADB (angle the cevian makes with BC)
• α = angle BAD
• β = angle CAD

Simple Analogy

Imagine cutting a pizza slice with a line. The m-n theorem tells you how the angles change based on where you make the cut!

Example

In triangle ABC, D divides BC such that BD:DC = 2:3 B = 50°, C = 70°

Find angle θ (angle ADB):

(2+3)·cot(θ) = 2·cot(50°) - 3·cot(70°)

5·cot(θ) = 2×0.839 - 3×0.364

5·cot(θ) = 1.678 - 1.092 = 0.586

cot(θ) = 0.117

θ = 83.3°

When to Use It

✅ Problems involving cevians (lines from vertex to opposite side) ✅ When a side is divided in a given ratio

6️⃣ Stewart’s Theorem: The Length Finder

What Is It?

This theorem finds the length of a cevian when you know the three sides and how the cevian divides the opposite side.

The Formula

If cevian AD has length d, and D divides BC into segments m and n:

b²·m + c²·n = a(d² + m·n)

Or in expanded form:

man + dad = bmb + cnc

(A fun memory trick: “A man and his dad visit bmb and cnc!”)

The Diagram

graph TD
    A((A)) -->|c| B((B))
    A -->|b| C((C))
    A -->|d| D((D))
    B -->|m| D
    D -->|n| C

Example

In triangle ABC:

• AB = c = 6
• AC = b = 8
• BC = a = 10
• D is the midpoint of BC (so m = n = 5)

Find cevian length d (the median):

b²·m + c²·n = a(d² + m·n)

64×5 + 36×5 = 10(d² + 25)

320 + 180 = 10d² + 250

500 = 10d² + 250

d² = 25

d = 5

The median from A has length 5!

When to Use It

✅ Finding the length of medians ✅ Finding the length of angle bisectors ✅ Any cevian length problem

7️⃣ Ptolemy’s Theorem: The Circle Secret

What Is It?

This beautiful theorem works for cyclic quadrilaterals (four points on a circle). It connects the diagonals to the sides!

The Formula

For a cyclic quadrilateral ABCD:

AC × BD = AB × CD + BC × AD

In words:

Product of diagonals = Sum of products of opposite sides

Simple Analogy

Imagine four friends holding hands in a circle. Ptolemy says: “The distance you’d walk diagonally equals a special sum of the side paths!”

The Diagram

graph TD
    A((A)) --- B((B))
    B --- C((C))
    C --- D((D))
    D --- A
    A -.- C
    B -.- D
    style A fill:#4ECDC4
    style B fill:#FF6B6B
    style C fill:#4ECDC4
    style D fill:#FF6B6B

Example

A cyclic quadrilateral has:

• AB = 5, BC = 6, CD = 7, DA = 8
• Diagonal BD = 10

Find diagonal AC:

AC × BD = AB × CD + BC × AD

AC × 10 = 5 × 7 + 6 × 8

10 × AC = 35 + 48

AC = 83/10 = 8.3

Special Case: Triangle in a Circle!

When D coincides with A (making a triangle ABC inscribed in a circle), Ptolemy’s theorem gives us the Sine Rule!

When to Use It

✅ Cyclic quadrilateral problems ✅ Proving the Sine Rule ✅ Finding diagonals when sides are known

🎯 Quick Reference: Which Formula When?

🌟 The Connection Web

All these formulas are connected! They’re different windows into the same beautiful world of triangles.

graph TD
    SR["Sine Rule"] --> NP["Napier's Analogy]
    SR --> MW[Mollweide]
    CR[Cosine Rule] --> PF[Projection Formula]
    CR --> LT[Law of Tangents]
    CR --> ST[Stewart's Theorem"]
    ST --> MN["m-n Theorem"]
    PT["Ptolemy"] --> SR
    style SR fill:#4ECDC4
    style CR fill:#FF6B6B
    style PT fill:#FFE66D

🚀 You’re Now a Triangle Master!

These seven formulas are your secret weapons. Each one solves problems that the basic rules can’t easily handle.

Remember:

• Projection = Shadows on sides
• Napier & Tangents = Angle-side ratios
• Mollweide = The ultimate checker
• m-n Theorem = Cevian angles
• Stewart = Cevian lengths
• Ptolemy = Circle magic

Go forth and conquer every triangle problem! 🔺✨