Law of Sines & Cosines

⚖️ 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

📝 Worked Examples

Example 1: AAS — Find a Missing Side

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

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

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

🧠 Step-by-Step Strategy

Solving AAS/ASA Triangles

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

Key Formula Rearrangements

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

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: $a = 8$, $b = 11$, $C = 40°$

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

Law of Sines Basics 🎯

Solve Triangles 🧮

Round to the nearest integer.

1) $\triangle ABC$: $A = 40°$, $B = 60°$, $a = 10$. Find $C$ in degrees. (e.g., if $A = 50°, B = 70°$, then $C = 180 - 50 - 70 = 60$)

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 $= \frac{1}{2}(8)(10)\sin 60° = 40(0.866) = 35$)

Matching 🔽

Exit Quiz ✅

⚠️ 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 $\sin B = \frac{b \sin A}{a}$.

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: $A = 30°$, $B \approx 131.4°$, $C \approx 18.6°$

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 $B_1$)
• Else: Two triangles (use both $B_1$ and $B_2$)

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 $B_2$? (e.g., if $B_1 = 45°$, then $B_2 = 180 - 45 = 135$)

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 🔽

Exit Quiz ✅

📐 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

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

Equivalently:

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

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!

When to Use It

📝 Worked Examples

Example 1: SAS — Find a Side

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

$c^2 = 7^2 + 10^2 - 2(7)(10)\cos 50°$ $c^2 = 49 + 100 - 140(0.6428) = 149 - 89.99 = 59.01$ $c = \sqrt{59.01} \approx 7.68$

Example 2: SSS — Find an Angle

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

Find angle $C$: $c^2 = a^2 + b^2 - 2ab\cos C$ $81 = 25 + 64 - 80\cos C$ $81 = 89 - 80\cos C$ $\cos C = \frac{89 - 81}{80} = \frac{8}{80} = 0.1$ $C = \arccos(0.1) \approx 84.3°$

Example 3: Verify with Pythagorean Triple

$a = 3$, $b = 4$, $c = 5$. Find $C$: $\cos C = \frac{9 + 16 - 25}{24} = \frac{0}{24} = 0 \implies C = 90°$ ✓

🔢 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:

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

Exit Quiz ✅

🔄 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

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

Mixed Strategy

Many problems require both laws:

1. Start with Law of Cosines to find a missing side or angle
2. 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 $A$. $\frac{\sin A}{9} = \frac{\sin 75°}{13.0} \implies \sin A = \frac{9(0.9659)}{13.0} \approx 0.6687$ $A \approx 41.9°$

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 $c$): $\cos C = \frac{36 + 64 - 121}{96} = \frac{-21}{96} \approx -0.219$ $C \approx 102.6°$

Step 2 (Law of Sines): $\sin A = \frac{6 \sin 102.6°}{11} \approx \frac{6(0.976)}{11} \approx 0.532$ $A \approx 32.2°$

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

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 $= \sqrt{15 \cdot 8 \cdot 5 \cdot 2}$. What is the area to the nearest integer? (e.g., $\sqrt{900} = 30$)

3) Triangle with $a = 10$, $b = 10$, included angle $C = 30°$. Area = ? (e.g., $\frac{1}{2}(8)(8)\sin 60° = 32(0.866) = 28$)

Method Selection 🔽

Exit Quiz ✅

🌍 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

The Triangulation Method

To find an inaccessible distance:

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

📝 Worked Examples

Example 1: Distance Across a River

From points $A$ and $B$ on one bank, $AB = 200$ m. Angles to a tree $C$ on the other bank: $\angle A = 72°$, $\angle B = 63°$.

$\angle C = 180° - 72° - 63° = 45°$

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

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 $180° - 60° = 120°$):

$R^2 = F_1^2 + F_2^2 - 2F_1 F_2 \cos 120°$ $R^2 = 900 + 1600 - 2400(-0.5) = 2500 + 1200 = 3700$ $R = \sqrt{3700} \approx 60.8 \text{ N}$

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

$R^2 = 30^2 + 40^2 + 2(30)(40)\cos 60° = 900 + 1600 + 1200 = 3700$ $R \approx 60.8 \text{ N}$

Direction of the Resultant

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

$\frac{\sin \alpha}{40} = \frac{\sin 60°}{60.8}$ $\sin \alpha = \frac{40(0.866)}{60.8} \approx 0.570 \implies \alpha \approx 34.7°$

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 $R$ to nearest integer. (e.g., $\sqrt{100 + 100 + 100} \approx 17$)

Application Matching 🔽

Exit Quiz ✅

📏 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

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

⭕ 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 = $60°$.

$d_1^2 = 36 + 100 - 120\cos 60° = 136 - 60 = 76 \implies d_1 = \sqrt{76} \approx 8.72$ $d_2^2 = 36 + 100 + 120\cos 60° = 136 + 60 = 196 \implies d_2 = 14$

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 $= 6$, $s = 6$: $r = \frac{6}{6} = 1$)

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 🔽

Exit Quiz ✅

🏆 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

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

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?

$a = 4$, $b = 5$, $c = 8$. $\cos C = \frac{16 + 25 - 64}{40} = \frac{-23}{40} = -0.575$. $C = \arccos(-0.575) \approx 125.1°$. 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: $d^2 = 25 + 64 - 80\cos 30° = 89 - 69.3 = 19.7$ $d \approx 4.44$ km.

Problem 4: Complete Solution

$a = 11$, $b = 14$, $C = 72°$. $c^2 = 121 + 196 - 308\cos 72° = 317 - 95.2 = 221.8$, $c \approx 14.9$. $\sin A = \frac{11\sin 72°}{14.9} \approx 0.702$, $A \approx 44.6°$. $B = 180° - 72° - 44.6° = 63.4°$.

⚠️ Common Exam Mistakes

Quick Formula Sheet

• Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
• 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): $\sqrt{s(s-a)(s-b)(s-c)}$

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., $\sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84$)

Master Review 🔽

Final Exit Quiz ✅