🎯⭐ INTERACTIVE LESSON

Law of Sines & Cosines

Learn step-by-step with interactive practice!

← Back to Standard Lesson

Law of Sines & Cosines - Complete Interactive Lesson

Part 1: Law of Sines

⚖️ Law of Sines

Part 1 of 7

The Law of Sines lets us solve oblique triangles (no right angle) when we know an angle–side pair.

The Law of Sines

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

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

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

When to Use It

Known	Case Name	Law of Sines?
Two angles + one side (AAS or ASA)	Angle–Angle–Side	✅ Yes
Two sides + angle opposite one (SSA)	Side–Side–Angle	✅ Yes (watch for ambiguous case!)
Two sides + included angle (SAS)	Side–Angle–Side	❌ Use Law of Cosines
Three sides (SSS)	Side-Side-Side	❌ Use Law of Cosines

📝 Worked Examples

Example 1: AAS — Find a Missing Side

In $\triangle ABC$ : $A = 42°$ , $B = 73°$ , $a = 12$ .

🧠 Step-by-Step Strategy

Solving AAS/ASA Triangles

Find the third angle using $A + B + C = 180°$
Set up the proportion using the known angle–side pair
Cross-multiply and solve

Key Formula Rearrangements

To find side $b$ : $b = \frac{a \sin B}{\sin A}$

Law of Sines Basics 🎯

Solve Triangles 🧮

Round to the nearest integer.

1) $\triangle ABC$ : $A = 40°$ , $B = 60°$ , $a = 10$ . Find in degrees. (e.g., if , then )

Matching 🔽

Exit Quiz ✅

Part 2: Ambiguous Case

⚠️ The Ambiguous Case (SSA)

Part 2 of 7

When given two sides and an angle opposite one of them (SSA), there might be zero, one, or two possible triangles.

Why It's Ambiguous

Given $a$ , $b$ , and angle $A$ :

We compute $\sin B = \frac{b \sin A}{a}$ .

Part 3: Law of Cosines

📐 Law of Cosines

Part 3 of 7

The Law of Cosines generalizes the Pythagorean theorem to any triangle — even those without a right angle.

The Law of Cosines

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

Part 4: Area of Triangles

🔄 Choosing the Right Law & Combined Problems

Part 4 of 7

Knowing when to use Law of Sines vs Law of Cosines is half the battle. This part helps you develop that judgment.

Decision Guide

$\boxed{\text{Do you have a complete angle–side pair (angle and its opposite side)?}}$

Answer	Case	Method
Yes + need another side/angle	AAS, ASA, SSA	Law of Sines
No + have SAS	Two sides + included angle	Law of Cosines (find side)

Part 5: Applications

🌍 Applications — Navigation & Surveying

Part 5 of 7

The Laws of Sines and Cosines solve real-world problems involving distances and angles that can't be measured directly.

Application Types

Application	Typical Setup
Navigation	Find distance between two points using bearings
Surveying	Measure inaccessible distances from two known positions
Engineering	Force resolution in non-rectangular systems
Aviation	Distance between waypoints on non-straight routes

The Triangulation Method

To find an inaccessible distance:

Measure a baseline (known distance between two observation points)
Measure the angles from each end of the baseline to the target
Use Law of Sines to compute the unknown distances

📝 Worked Examples

Example 1: Distance Across a River

From points $A$ and on one bank, m. Angles to a tree on the other bank: , .

Part 6: Problem-Solving Workshop

📏 Advanced Problem-Solving Techniques

Part 6 of 7

This part covers problems that require creative setups, multi-step approaches, and combining both laws.

Advanced Techniques

Breaking complex shapes into triangles — add diagonals
Using supplementary angles in parallelograms
Computing distances in 3D by projecting onto 2D triangles
Inscribed circle radius: $r = \frac{\text{Area}}{s}$
Circumscribed circle radius:

Part 7: Review & Applications

🏆 Law of Sines & Cosines — Full Synthesis

Part 7 of 7

Master review covering all concepts: Law of Sines (including ambiguous case), Law of Cosines, area formulas, and applications.

Complete Decision Tree

$\boxed{\text{Given info} \to \text{Identify case} \to \text{Choose law} \to \text{Solve}}$

Case

Step 1: Find $C$ : $C = 180° - 42° - 73° = 65°$

Step 2: Use Law of Sines to find $b$ : $\frac{a}{\sin A} = \frac{b}{\sin B}$ $\frac{12}{\sin 42°} = \frac{b}{\sin 73°}$ $b = \frac{12 \sin 73°}{\sin 42°} = \frac{12(0.9563)}{0.6691} \approx 17.15$

Example 2: ASA — Find a Missing Side

In $\triangle ABC$ : $A = 50°$ , $C = 60°$ , $b = 20$ .

$B = 180° - 50° - 60° = 70°$

$\frac{b}{\sin B} = \frac{a}{\sin A} \implies a = \frac{20 \sin 50°}{\sin 70°} \approx \frac{20(0.766)}{0.940} \approx 16.30$

Example 3: Finding an Angle

In $\triangle ABC$ : $a = 10$ , $b = 15$ , $A = 30°$ .

$\frac{\sin B}{b} = \frac{\sin A}{a} \implies \sin B = \frac{15 \sin 30°}{10} = \frac{15(0.5)}{10} = 0.75$ $B = \arcsin(0.75) \approx 48.6°$

b = \frac{a s i n B}{s i n A}

To find angle $B$ : $\sin B = \frac{b \sin A}{a}$ , then $B = \arcsin(\ldots)$

Triangle Area with Law of Sines

The area of any triangle can be found using: $\boxed{\text{Area} = \frac{1}{2}ab\sin C}$

This formula uses two sides and their included angle.

Example: $a = 8$ , $b = 11$ , $C = 40°$

$\text{Area} = \frac{1}{2}(8)(11)\sin 40° = 44(0.6428) \approx 28.3 \text{ sq units}$

A = 50°, B = 70°

C = 180 - 50 - 70 = 60

2) Same triangle: find $b$ to nearest integer. Use $b = \frac{a \sin B}{\sin A}$ . (e.g., $\frac{10 \cdot \sin 45°}{\sin 30°} = \frac{10(0.707)}{0.5} = 14$ )

3) Area of triangle with $a = 10$ , $b = 14$ , included angle $= 50°$ . Round to nearest integer. (e.g., Area $= \frac{1}{2}(8)(10)\sin 60° = 40(0.866) = 35$ )

Condition	# of Triangles
$\sin B > 1$	0 triangles (impossible)
$\sin B = 1$	1 triangle ( $B = 90°$ )
$\sin B < 1$ and $A$ is obtuse	1 triangle (only acute $B$ works)
$\sin B < 1$ and $A$ is acute	1 or 2 triangles — check both $B$ and $180° - B$

The Key Test for Two Triangles

If $\sin B < 1$ and angle $A$ is acute, compute:

$B_1 = \arcsin(\frac{b \sin A}{a})$ and $B_2 = 180° - B_1$

If $A + B_2 < 180°$ : TWO triangles exist
If $A + B_2 \geq 180°$ : ONE triangle (only $B_1$ works)

📝 Worked Examples

Example 1: Two Triangles

Given: $a = 8$ , $b = 12$ , $A = 30°$

$\sin B = \frac{12 \sin 30°}{8} = \frac{12(0.5)}{8} = 0.75$

$B_1 = \arcsin(0.75) \approx 48.6°$ , $B_2 = 180° - 48.6° = 131.4°$

Check: $A + B_2 = 30° + 131.4° = 161.4° < 180°$ ✓

Triangle 1: $A = 30°$ , $B \approx 48.6°$ , $C \approx 101.4°$ Triangle 2: $A = 30°$ , ,

Example 2: One Triangle

Given: $a = 15$ , $b = 10$ , $A = 60°$

$\sin B = \frac{10 \sin 60°}{15} = \frac{10(0.866)}{15} \approx 0.577$

$B_1 \approx 35.3°$ , $B_2 = 180° - 35.3° = 144.7°$

Check: $A + B_2 = 60° + 144.7° = 204.7° > 180°$ ✗

Only one triangle: $B \approx 35.3°$ , $C \approx 84.7°$ .

Example 3: No Triangle

Given: $a = 5$ , $b = 20$ , $A = 40°$

$\sin B = \frac{20 \sin 40°}{5} = \frac{20(0.643)}{5} \approx 2.57 > 1$

No triangle exists — impossible!

🔄 Decision Flowchart for SSA

$\boxed{\text{Step 1: Compute } \sin B = \frac{b \sin A}{a}}$

If $\sin B > 1$ : No triangle. Stop.

If $\sin B = 1$ : One right triangle. $B = 90°$ .

If $\sin B < 1$ :

$B_1 = \arcsin(\sin B)$

$B_2 = 180° - B_1$

If $A + B_1 \geq 180°$ : No triangle
Else if $A + B_2 \geq 180°$ : One triangle (use )

Quick Rule of Thumb

If $a \geq b$ (the side opposite the given angle is longer), there's always exactly one triangle. The ambiguous case only arises when $a < b$ .

Ambiguous Case Quiz 🎯

Ambiguous Case Calculations 🧮

1) Given $a = 10$ , $b = 14$ , $A = 35°$ . Compute $\sin B$ to 2 decimal places. (e.g., $\frac{12 \sin 30°}{8} = \frac{12(0.5)}{8} = 0.75$ )

2) In a triangle with $\sin B = 0.8$ and this might be ambiguous: $B_1 = \arcsin(0.8) \approx 53°$ , what is ? (e.g., if , then )

3) How many triangles exist if $a = 20$ , $b = 10$ , $A = 50°$ ? (When $a > b$ , the side opposite the known angle is longer.) Answer: 0, 1, or 2.

Classify the Case 🔽

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

Connection to the Pythagorean Theorem

When $C = 90°$ : $\cos 90° = 0$ , so $c^2 = a^2 + b^2 - 0 = a^2 + b^2$ ← the Pythagorean theorem!

Given Information	Use Law of Cosines?
SAS (two sides + included angle)	✅ Find the third side
SSS (three sides)	✅ Find any angle
AAS or ASA	❌ Use Law of Sines

📝 Worked Examples

Example 1: SAS — Find a Side

In $\triangle ABC$ : $a = 7$ , $b = 10$ , $C = 50°$ .

$c^2 = 7^2 + 10^2 - 2(7)(10)\cos 50°$

Example 2: SSS — Find an Angle

In $\triangle ABC$ : $a = 5$ , $b = 8$ , $c = 9$ .

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

Example 3: Verify with Pythagorean Triple

$a = 3$ , $b = 4$ , $c = 5$ . Find $C$ : $\cos C = \frac{}{}$ ✓

🔢 Rearranging for Angles

The Angle Formula

To find angle $C$ directly:

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

Similarly: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}, \qquad \cos B = \frac{a^2 + c^2 - b^2}{2ac}$

Checking Triangle Type

Using the Law of Cosines, we can determine the triangle type:

If $\cos C > 0$	$C < 90°$	Acute triangle (if all angles are acute)
If $\cos C = 0$

Strategy: Which Angle to Find First?

When given SSS, find the largest angle first (opposite the longest side). This avoids ambiguity because $\arccos$ always gives a unique answer.

Law of Cosines Quiz 🎯

Compute with Law of Cosines 🧮

1) $a = 3$ , $b = 5$ , $C = 120°$ . Find $c^2$ . (e.g., $c^2 = 6^2 + 8^2 - 2(6)(8)\cos 60° = 100 - 48 = 52$ )

2) $a = 8$ , $b = 6$ , $c = 10$ . Find $\cos C$ as a fraction in the form p/q. (e.g., , write 8/80)

3) In the triangle from #2, is $C$ acute, right, or obtuse? Write: acute, right, or obtuse. (e.g., $\cos C = 0.5 > 0$ means acute)

Identify the Approach 🔽

Many problems require both laws:

Start with Law of Cosines to find a missing side or angle
Switch to Law of Sines (which is easier) for the remaining parts

📝 Worked Examples

Example 1: SAS → Find All Parts

$a = 9$ , $b = 12$ , $C = 75°$ .

Step 1 (Law of Cosines): Find $c$ . $c^2 = 81 + 144 - 216\cos 75° = 225 - 216(0.2588) = 225 - 55.9 = 169.1$ $c \approx 13.0$

Step 2 (Law of Sines): Find $A$ . $\frac{\sin A}{9} = \frac{\sin 75°}{13.0} \implies \sin A = \frac{9(0.9659)}{13.0} \approx 0.6687$

Step 3: $B = 180° - 75° - 41.9° = 63.1°$

Example 2: SSS → Find All Angles

$a = 6$ , $b = 8$ , $c = 11$ .

Step 1: Find the largest angle (opposite longest side $c$ ): $\cos C = \frac{36 + 64 - 121}{96} = \frac{-21}{96} \approx -0.219$

Step 2 (Law of Sines): $\sin A = \frac{6 \sin 102.6°}{11} \approx \frac{6(0.976)}{11} \approx 0.532$

Step 3: $B = 180° - 102.6° - 32.2° = 45.2°$

📐 Hero's Formula (Heron's Formula)

When you know all three sides, you can find the area directly:

$\boxed{\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}}$

where $s = \frac{a + b + c}{2}$ is the semi-perimeter.

Example: $a = 5$ , $b = 12$ , $c = 13$

$s = \frac{5 + 12 + 13}{2} = 15$

$\text{Area} = \sqrt{15(15-5)(15-12)(15-13)} = \sqrt{15 \cdot 10 \cdot 3 \cdot 2} = \sqrt{900} = 30$

Verify: This is a 5-12-13 right triangle. Area $= \frac{1}{2}(5)(12) = 30$ ✓

When to Use Each Area Formula

Formula	When to Use
$\frac{1}{2}bh$	Base and height known
$\frac{1}{2}ab\sin C$

Strategy Quiz 🎯

Area Calculations 🧮

1) $a = 7$ , $b = 10$ , $c = 13$ . Compute $s$ (semi-perimeter). (e.g., for sides 3, 4, 5: $s = \frac{3+4+5}{2} = 6$ )

2) Using $s = 15$ from the triangle with sides $a = 7$ , $b = 10$ , $c = 13$ : Area . What is the area to the nearest integer? (e.g., )

3) Triangle with $a = 10$ , $b = 10$ , included angle $C = 30°$ . Area = ? (e.g., $\frac{1}{2} (8) (8) \sin 60 ° = 32 (0.866)$ )

Method Selection 🔽

$\angle C = 180° - 72° - 63° = 45°$

$\frac{AC}{\sin B} = \frac{AB}{\sin C}$ $AC = \frac{200 \sin 63°}{\sin 45°} = \frac{200(0.891)}{0.707} \approx 252.1 \text{ m}$

Example 2: Two Ships from a Lighthouse

A lighthouse sees Ship 1 at bearing N40°E, distance 8 km, and Ship 2 at bearing S50°E, distance 6 km.

The angle at the lighthouse = $40° + 50° = 90°$ .

By Law of Cosines: $d^2 = 8^2 + 6^2 - 2(8)(6)\cos 90° = 64 + 36 - 0 = 100$ $d = 10 \text{ km}$

Example 3: Hiking Problem

A hiker walks 5 km on bearing 060°, then turns and walks 7 km on bearing 150°. How far from the start?

Angle between paths = $150° - 60° = 90°$ (the turn angle is $180° - 90° = 90°$ ).

Actually, the angle in the triangle at the turning point = $180° - (150° - 60°) = 90°$ .

$d = \sqrt{5^2 + 7^2} = \sqrt{74} \approx 8.60 \text{ km}$

⚡ Force and Velocity Problems

Resultant of Two Forces

Two forces $F_1 = 30$ N and $F_2 = 40$ N act at an angle of $60°$ to each other.

The resultant magnitude uses the Law of Cosines (the angle in the triangle is $180° - 60° = 120°$ ):

$R^2 = F_1^2 + F_2^2 - 2F_1 F_2 \cos 120°$

Wait — actually for the parallelogram law, the angle between the forces in the triangle is the supplement:

$R^2 = 30^2 + 40^2 + 2(30)(40)\cos 60° = 900 + 1600 + 1200 = 3700$

Direction of the Resultant

Use Law of Sines to find the angle $\alpha$ the resultant makes with $F_1$ :

$\frac{\sin \alpha}{40} = \frac{\sin 60°}{60.8}$

Applications Quiz 🎯

Solve Applications 🧮

Round to nearest integer.

1) Baseline $AB = 100$ m. Angles to target $C$ : $\angle A = 70°$ , $\angle B = 65°$ . Find $\angle C$ in degrees. (e.g., $\angle C = 180 - 72 - 63 = 45$ )

2) Two ships 12 km apart at an angle of 90° from a port. Find the distance between the ships. (e.g., $\sqrt{8^2 + 6^2} = \sqrt{100} = 10$ )

3) Forces of 50 N and 50 N at 60° to each other. Resultant $R = \sqrt{50^2 + 50^2 + 2(50)(50)\cos 60°}$ . Find to nearest integer. (e.g., )

Application Matching 🔽

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

R = \frac{a}{2 s i n A}

⭕ The Circumscribed Circle

Circumradius Formula

For any triangle with circumradius $R$ :

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

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

Example 1: Find the Circumradius

$\triangle ABC$ : $a = 10$ , $A = 30°$ .

$R = \frac{10}{2\sin 30°} = \frac{10}{2(0.5)} = \frac{10}{1} = 10$

Inradius Formula

$\boxed{r = \frac{\text{Area}}{s}}$

where $s$ is the semi-perimeter.

Example 2: Find the Inradius

$a = 3$ , $b = 4$ , $c = 5$ (right triangle).

Area = $\frac{1}{2}(3)(4) = 6$ . $s = \frac{12}{2} = 6$ .

$r = \frac{6}{6} = 1$

◆ Parallelogram Diagonals

Finding Diagonal Lengths

A parallelogram with sides $a$ and $b$ and angle $\theta$ has diagonals:

$d_1^2 = a^2 + b^2 - 2ab\cos\theta$ $d_2^2 = a^2 + b^2 + 2ab\cos\theta$

(One diagonal subtends $\theta$ , the other subtends $180° - \theta$ .)

Example 3: Parallelogram ABCD

Sides $a = 6$ , $b = 10$ , angle = $60°$ .

$d_1^2 = 36 + 100 - 120\cos 60° = 136 - 60 = 76 \implies d_1 = \sqrt{76} \approx 8.72$

Property Check

$d_1^2 + d_2^2 = 76 + 196 = 272 = 2(36 + 100) = 2(a^2 + b^2)$

This confirms: In any parallelogram, the sum of the squares of the diagonals equals twice the sum of the squares of the sides.

Advanced Quiz 🎯

Advanced Calculations 🧮

1) $\triangle ABC$ : $c = 14$ , $C = 90°$ . Find circumradius $R$ . (e.g., $R = \frac{a}{2\sin A}$ ; for $a=10$ , $A=30°$ : $R = \frac{10}{2(0.5)} = 10$ )

2) Triangle with area $= 24$ and semi-perimeter $s = 8$ . Find inradius $r$ . (e.g., area $= 6$ , $s = 6$ : )

3) Parallelogram with sides 5 and 12, angle $60°$ . Find the shorter diagonal to nearest integer. Use $d^2 = 25 + 144 - 120\cos 60°$ . (e.g., $\sqrt{76} \approx 9$ )

Advanced Matching 🔽

Complete Area Formulas

$\text{Area} = \frac{1}{2}bh = \frac{1}{2}ab\sin C = \sqrt{s(s-a)(s-b)(s-c)}$

Complete Circle Formulas

$R = \frac{a}{2\sin A}, \qquad r = \frac{\text{Area}}{s}$

📝 Mixed Review Problems

Problem 1: Identify and Solve (AAS)

$A = 48°$ , $B = 67°$ , $a = 15$ . $C = 180° - 48° - 67° = 65°$ . $b = \frac{15\sin 67°}{\sin 48°} = \frac{15(0.921)}{0.743} \approx 18.6$ .

Problem 2: SSS — Is it Obtuse?

$a = 4$ , $b = 5$ , $c = 8$ . $\cos C = \frac{16 + 25 - 64}{40} = \frac{-23}{40} = -0.575$ . . Yes, it's obtuse.

Problem 3: SAS Application

Two roads diverge at $30°$ . After 5 km on one road and 8 km on the other, the distance between endpoints: $d^2 = 25 + 64 - 80\cos 30° = 89 - 69.3 = 19.7$ km.

Problem 4: Complete Solution

$a = 11$ , $b = 14$ , $C = 72°$ . $c^2 = 121 + 196 - 308\cos 72° = 317 - 95.2 = 221.8$ , . , . .

⚠️ Common Exam Mistakes

Mistake	Correction
Using Law of Sines for SAS	No complete pair exists — use Law of Cosines first
Forgetting the ambiguous case in SSA	Always check if $\sin B < 1$ allows two values
Wrong angle in area formula	$\frac{1}{2}ab\sin C$ — $C$ must be the included angle between $a$ and $b$
Confusing supplement: $\cos(180°-\theta) = -\cos\theta$	In force problems, the triangle angle is the supplement of the physical angle
Rounding too early	Keep at least 4 decimal places in intermediate steps

Quick Formula Sheet

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

Comprehensive Quiz 🎯

Final Calculations 🧮

1) Triangle with $a = 5$ , $b = 12$ , $c = 13$ . Is the largest angle 90°? Enter the value of $\cos C$ (where $C$ is opposite side 13). (e.g., $\cos C = \frac{49+64-169}{112} = \frac{-56}{112}$ , write $-0.5$ )

2) Circumradius of a triangle where $a = 10$ and $A = 30°$ . (e.g., $R = \frac{8}{2\sin 45°} \approx 5.66$ )

3) Area using Heron's formula: $a = 5$ , $b = 12$ , $c = 13$ . $s = 15$ . Area = ? (e.g., )

Master Review 🔽

Final Exit Quiz ✅

Else: Two triangles (use both

B_1

and

B_2

)

B_2 = 180 - 45 = 135

c^2 = 49 + 100 - 140(0.6428) = 149 - 89.99 = 59.01

c = \sqrt{59.01} \approx 7.68

81 = 25 + 64 - 80\cos C

81 = 89 - 80\cos C

\cos C = \frac{89 - 81}{80} = \frac{8}{80} = 0.1

C = \arccos(0.1) \approx 84.3°

cos C = \frac{9 + 16 - 25}{24} = \frac{0}{24} = 0 ⟹ C = 90°

cos C = \frac{25 + 64 - 81}{2 ( 5 ) ( 8 )} = \frac{8}{80}

= \sqrt{15 \cdot 8 \cdot 5 \cdot 2}

\frac{1}{2} (8) (8) sin 60° = 32 (0.866) = 28

R^2 = 900 + 1600 - 2400(-0.5) = 2500 + 1200 = 3700

R = \sqrt{3700} \approx 60.8 \text{ N}

R \approx 60.8 \text{ N}

\sin \alpha = \frac{40(0.866)}{60.8} \approx 0.570 \implies \alpha \approx 34.7°

\sqrt{100 + 100 + 100} \approx 17

d_2^2 = 36 + 100 + 120\cos 60° = 136 + 60 = 196 \implies d_2 = 14

r = \frac{6}{6} = 1

C = \arccos(-0.575) \approx 125.1°

\sin A = \frac{11\sin 72°}{14.9} \approx 0.702

B = 180° - 72° - 44.6° = 63.4°

Law of Cosines:

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

Angle from SSS:

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

Area (SAS):

\frac{1}{2}ab\sin C

Area (SSS):

\sqrt{s(s-a)(s-b)(s-c)}

\sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84

AAS/ASA	Sines	Find 3rd angle, then use proportions
SSA	Sines	Check ambiguous case first
SAS	Cosines	Find opposite side, then switch to Sines
SSS	Cosines	Find largest angle (opposite longest side)

Law of Sines & Cosines

Law of Sines & Cosines - Complete Interactive Lesson

Part 1: Law of Sines

⚖️ Law of Sines

The Law of Sines

When to Use It

📝 Worked Examples

Example 1: AAS — Find a Missing Side

🧠 Step-by-Step Strategy

Solving AAS/ASA Triangles

Key Formula Rearrangements

Part 2: Ambiguous Case

⚠️ The Ambiguous Case (SSA)

Why It's Ambiguous

Part 3: Law of Cosines

📐 Law of Cosines

The Law of Cosines

Part 4: Area of Triangles

🔄 Choosing the Right Law & Combined Problems

Decision Guide

Part 5: Applications

🌍 Applications — Navigation & Surveying

Application Types

The Triangulation Method

📝 Worked Examples

Example 1: Distance Across a River

Part 6: Problem-Solving Workshop

📏 Advanced Problem-Solving Techniques

Advanced Techniques

Part 7: Review & Applications

🏆 Law of Sines & Cosines — Full Synthesis

Complete Decision Tree

Example 2: ASA — Find a Missing Side

Example 3: Finding an Angle

Triangle Area with Law of Sines

The Key Test for Two Triangles

📝 Worked Examples

Example 1: Two Triangles

Example 2: One Triangle

Example 3: No Triangle

🔄 Decision Flowchart for SSA

Quick Rule of Thumb

Connection to the Pythagorean Theorem

When to Use It

📝 Worked Examples

Example 1: SAS — Find a Side

Example 2: SSS — Find an Angle

Example 3: Verify with Pythagorean Triple

🔢 Rearranging for Angles

The Angle Formula

Checking Triangle Type

Strategy: Which Angle to Find First?

Mixed Strategy

📝 Worked Examples

Example 1: SAS → Find All Parts

Example 2: SSS → Find All Angles

📐 Hero's Formula (Heron's Formula)

Example: a=5a = 5a=5, b=12b = 12b=12, c=13c = 13c=13

When to Use Each Area Formula

Example 2: Two Ships from a Lighthouse

Example 3: Hiking Problem

⚡ Force and Velocity Problems

Resultant of Two Forces

Direction of the Resultant

⭕ The Circumscribed Circle

Circumradius Formula

Example 1: Find the Circumradius

Inradius Formula

Example 2: Find the Inradius

◆ Parallelogram Diagonals

Finding Diagonal Lengths

Example 3: Parallelogram ABCD

Property Check

Complete Area Formulas

Complete Circle Formulas

📝 Mixed Review Problems

Problem 1: Identify and Solve (AAS)

Problem 2: SSS — Is it Obtuse?

Problem 3: SAS Application

Problem 4: Complete Solution

Example: $a = 5$ , $b = 12$ , $c = 13$