Laws of Sines and Cosines

The Secret Tools for Unlocking Any Triangle 🔺

Imagine you’re a detective. Someone hands you a triangle and says, “Find the missing pieces!” But here’s the catch—you only have some clues, not all. How do you solve the puzzle?

Welcome to the Law of Sines and Law of Cosines—your two ultimate detective tools for cracking any triangle mystery!

The Big Picture: Why Do We Need These Laws?

Think of triangles like puzzles with 6 pieces: 3 sides and 3 angles. Usually, if someone gives you 3 pieces (with at least one side), you can figure out the rest!

But here’s the problem:

• Basic trigonometry (SOH-CAH-TOA) only works for right triangles (triangles with a 90° corner)
• Most triangles in the real world don’t have a 90° corner!

So we need new tools—and that’s where our heroes come in:

• Law of Sines: Works great when you have angle-side pairs
• Law of Cosines: Works great when you have three sides or two sides with the angle between them

Part 1: The Law of Sines 🎵

The Story Behind It

Picture a triangle sitting inside a perfect circle (we call this the circumscribed circle). Every triangle has one special circle that touches all three corners!

Here’s the magical discovery: no matter which side and opposite angle you pick, when you divide the side by the sine of its opposite angle, you always get the same number—and that number equals twice the radius of the special circle!

The Formula

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

Where:

• a, b, c = the three sides
• A, B, C = the angles opposite to those sides
• R = radius of the circumscribed circle

Deriving It (Where Does It Come From?)

Let’s prove this step by step:

Step 1: Draw a triangle ABC and its circumscribed circle with center O and radius R.

Step 2: Draw a diameter from vertex A through the center O to point D on the circle.

Step 3: Look at angle ADB. Because AD is a diameter, angle ADB = 90° (an angle in a semicircle is always 90°!)

Step 4: Notice that angle D and angle C both “look at” the same arc AB. In a circle, angles looking at the same arc are equal! So angle D = angle C.

Step 5: In the right triangle ABD:

• sin(D) = AB/AD = c/(2R)
• Since angle D = angle C: sin© = c/(2R)
• Rearranging: c/sin© = 2R

Step 6: The same logic works for any side-angle pair. Therefore: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$

The magic is proven! ✨

Part 2: Using the Law of Sines 🛠️

When to Use It

The Law of Sines is your best friend when you have:

• AAS (two angles + one side)
• ASA (two angles with the side between them)
• SSA (two sides + an angle opposite one of them) ⚠️ This one’s tricky!

Example 1: Finding a Missing Side (AAS)

Problem: In triangle ABC, angle A = 40°, angle B = 60°, and side a = 10 cm. Find side b.

Solution:

First, use Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B}$

$\frac{10}{\sin 40°} = \frac{b}{\sin 60°}$

Solve for b: $b = \frac{10 \times \sin 60°}{\sin 40°}$

$b = \frac{10 \times 0.866}{0.643} \approx 13.47 \text{ cm}$

Example 2: Finding a Missing Angle

Problem: In triangle XYZ, side x = 8, side y = 12, angle X = 35°. Find angle Y.

Solution: $\frac{x}{\sin X} = \frac{y}{\sin Y}$

$\frac{8}{\sin 35°} = \frac{12}{\sin Y}$

$\sin Y = \frac{12 \times \sin 35°}{8} = \frac{12 \times 0.574}{8} \approx 0.861$

$Y = \sin^{-1}(0.861) \approx 59.4°$

Part 3: The Ambiguous Case (SSA) ⚠️

Why SSA is Tricky

When you have two sides and an angle opposite one of them, something weird can happen: you might get zero, one, or TWO valid triangles!

Think of it like this: imagine you have a stick (side b) attached at a corner with a fixed angle (A). You swing the stick around—sometimes it hits the opposite side once, sometimes twice, sometimes not at all!

The SSA Decision Tree

Given: angle A, side a (opposite to A), and side b

Calculate h = b × sin(A) (this is the “height” from the angle)

Then check:

If A is ACUTE (< 90°):
├── If a < h → NO triangle exists
├── If a = h → ONE right triangle
├── If h < a < b → TWO triangles exist!
└── If a ≥ b → ONE triangle exists

If A is OBTUSE (> 90°):
├── If a ≤ b → NO triangle exists
└── If a > b → ONE triangle exists

Example: The Two-Triangle Case

Problem: Angle A = 40°, a = 10, b = 14. How many triangles?

Solution:

Calculate h = b × sin(A) = 14 × sin(40°) = 14 × 0.643 = 9.0

Since h (9.0) < a (10) < b (14), and A is acute: TWO triangles exist!

Finding both: $\sin B = \frac{b \times \sin A}{a} = \frac{14 \times 0.643}{10} = 0.900$

$B_1 = \sin^{-1}(0.900) = 64.2°$ $B_2 = 180° - 64.2° = 115.8°$

Both angles give valid triangles since both leave room for angle C to be positive!

Part 4: The Sine Rule with Circumradius 🔵

The Special Connection

Remember that 2R in our formula? It’s super useful!

$\frac{a}{\sin A} = 2R$

This means: Any side = 2R × sine of its opposite angle

Or flipped: R = a / (2 × sin A)

Example: Finding the Circumradius

Problem: In triangle PQR, side p = 15 cm and angle P = 50°. Find R.

Solution: $R = \frac{p}{2 \sin P} = \frac{15}{2 \times \sin 50°} = \frac{15}{2 \times 0.766} = \frac{15}{1.532} \approx 9.79 \text{ cm}$

Part 5: The Law of Cosines 📐

The Story Behind It

The Law of Cosines is like the Pythagorean Theorem’s older, wiser sibling. Remember a² + b² = c² for right triangles? Well, the Law of Cosines works for ALL triangles—it just adds a correction term!

The Formula

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

Or written for each side:

• $a^2 = b^2 + c^2 - 2bc \cos A$
• $b^2 = a^2 + c^2 - 2ac \cos B$
• $c^2 = a^2 + b^2 - 2ab \cos C$

Deriving It (Where Does It Come From?)

Step 1: Place triangle ABC on a coordinate plane with:

• Vertex C at origin (0, 0)
• Side a along the positive x-axis
• Vertex B at (a, 0)
• Vertex A at (b cos C, b sin C)

Step 2: Use the distance formula to find side c (distance from A to B): $c^2 = (b\cos C - a)^2 + (b\sin C - 0)^2$

Step 3: Expand: $c^2 = b^2\cos^2 C - 2ab\cos C + a^2 + b^2\sin^2 C$

Step 4: Group the b² terms: $c^2 = a^2 + b^2(\cos^2 C + \sin^2 C) - 2ab\cos C$

Step 5: Since cos²C + sin²C = 1: $c^2 = a^2 + b^2 - 2ab\cos C$

Bonus insight: When C = 90°, cos(90°) = 0, so the formula becomes c² = a² + b²—the Pythagorean Theorem! The Law of Cosines is the general version!

Part 6: Using the Law of Cosines 🔧

When to Use It

The Law of Cosines shines when you have:

• SAS (two sides + the angle between them)
• SSS (all three sides, need to find angles)

Example 1: Finding a Side (SAS)

Problem: In triangle ABC, b = 7, c = 9, angle A = 52°. Find side a.

Solution: $a^2 = b^2 + c^2 - 2bc \cos A$ $a^2 = 7^2 + 9^2 - 2(7)(9)\cos 52°$ $a^2 = 49 + 81 - 126 \times 0.616$ $a^2 = 130 - 77.6 = 52.4$ $a = \sqrt{52.4} \approx 7.24$

Example 2: Finding an Angle (SSS)

Problem: In a triangle, a = 8, b = 6, c = 10. Find angle C.

Solution:

Rearrange the formula to solve for cos C: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ $\cos C = \frac{8^2 + 6^2 - 10^2}{2(8)(6)}$ $\cos C = \frac{64 + 36 - 100}{96} = \frac{0}{96} = 0$ $C = \cos^{-1}(0) = 90°$

Wow! This is actually a right triangle! (6-8-10 is a Pythagorean triple)

Part 7: Choosing the Right Tool 🎯

Here’s your quick decision guide:

graph TD
    A["What information do you have?"] --> B{Three sides?}
    B -->|Yes SSS| C["Use Law of COSINES"]
    B -->|No| D{Two sides + included angle?}
    D -->|Yes SAS| C
    D -->|No| E{Angle-Side-Angle or<br>Angle-Angle-Side?}
    E -->|Yes ASA/AAS| F["Use Law of SINES"]
    E -->|No| G{Two sides + non-included angle?}
    G -->|Yes SSA| H["Use Law of SINES&lt;br&gt;Watch for ambiguous case!"]

Summary Table

Quick Reference Card 📋

Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$

Law of Cosines: $c^2 = a^2 + b^2 - 2ab\cos C$

Circumradius: $R = \frac{a}{2\sin A}$

Ambiguous Case (SSA): Calculate h = b sin A

• a < h → 0 triangles
• a = h → 1 right triangle
• h < a < b (acute A) → 2 triangles
• a ≥ b → 1 triangle

You’ve Got This! 🎉

You now have two powerful tools in your geometry toolkit:

• Law of Sines for angle-side pairs (with a bonus circumradius connection!)
• Law of Cosines for when you have adjacent pieces

Remember: every triangle puzzle can be solved. You just need to pick the right tool and watch out for that sneaky ambiguous case!

Now go crack some triangle mysteries! 🔍🔺