Sum of angles: $A + B + C = 180°$ (or $\pi$ radians)

Law of Sines

Formula

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

Or equivalently:

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

When to Use

Use the Law of Sines when you have:

1. AAS (Angle-Angle-Side): Two angles and one side
2. ASA (Angle-Side-Angle): Two angles and the included side
3. SSA (Side-Side-Angle): Two sides and a non-included angle ⚠️ Ambiguous case

Solving with Law of Sines

Given: Two angles and one side (AAS or ASA)

Steps:

1. Find the third angle: $C = 180° - A - B$
2. Use the law of sines to find the unknown sides
3. Use the ratio $\frac{a}{\sin A} = \frac{b}{\sin B}$

The Ambiguous Case (SSA)

When given two sides and a non-included angle (SSA), there may be:

• 0 solutions (no triangle exists)
• 1 solution (one unique triangle)
• 2 solutions (two different triangles)

Why it's ambiguous: The side opposite the known angle might "swing" to create two different triangles.

To determine the number of solutions:

Given sides $a$, $b$ and angle $A$ (where $a$ is opposite $A$):

1. If $a < b\sin A$: No triangle (side too short)
2. If $a = b\sin A$: One triangle (right triangle)
3. If $b\sin A < a < b$: Two triangles (ambiguous case)
4. If $a \geq b$: One triangle

Law of Cosines

Formulas

For any triangle:

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

Note: When $C = 90°$, $\cos C = 0$, and this reduces to the Pythagorean theorem: $c^2 = a^2 + b^2$

Solving for an Angle

Rearrange to solve for the cosine:

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

Then use $A = \arccos(\frac{b^2 + c^2 - a^2}{2bc})$

When to Use

Use the Law of Cosines when you have:

1. SAS (Side-Angle-Side): Two sides and the included angle
2. SSS (Side-Side-Side): All three sides

Strategy for Solving Triangles

Given Information → Method

Area of a Triangle

Using the Law of Sines, we can derive:

$\text{Area} = \frac{1}{2}ab\sin C = \frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B$

This is useful when you know two sides and the included angle.

Common Applications

1. Navigation: Finding distances and bearings
2. Surveying: Measuring inaccessible distances
3. Engineering: Analyzing forces in structures
4. Physics: Resolving vector components

Tips for Success

1. Draw a diagram and label all known values
2. Identify the given information (AAS, SAS, SSS, etc.)
3. Choose the appropriate law (Sines or Cosines)
4. Check your answer using the angle sum ($A + B + C = 180°$)
5. Watch for the ambiguous case with SSA
6. Use calculator in correct mode (degrees or radians)

$\frac{12}{\sin 35°} = \frac{15}{\sin B}$

$\sin B = \frac{15 \sin 35°}{12} = \frac{15(0.5736)}{12} = \frac{8.604}{12} = 0.717$

$B = \sin^{-1}(0.717) \approx 45.8°$

Note: There could be another solution $B' = 180° - 45.8° = 134.2°$, but we need to check if it's valid.

If $B = 134.2°$, then $A + B = 35° + 134.2° = 169.2° < 180°$, so this is also possible.

However, since $b > a$ and angle $B$ is opposite the larger side, we expect $B > A$.

Both solutions satisfy this, so we have the ambiguous case. Let's take $B \approx 45.8°$ as the acute solution.

Part (b): $C = 180° - A - B = 180° - 35° - 45.8° = 99.2°$

Part (c): Using Law of Sines:

$\frac{c}{\sin C} = \frac{a}{\sin A}$

$c = \frac{a \sin C}{\sin A} = \frac{12 \sin 99.2°}{\sin 35°} = \frac{12(0.9877)}{0.5736} \approx 20.7$

Step 1: Identify the case

This is SAS (two sides and the included angle), so use the Law of Cosines.

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

Substitute: $c^2 = 8^2 + 5^2 - 2(8)(5)\cos 60°$

Step 3: Calculate

$c^2 = 64 + 25 - 80 \cdot \frac{1}{2}$ $c^2 = 64 + 25 - 40$ $c^2 = 49$ $c = 7$

Answer: $c = 7$

Bonus: Find angle $A$

Now use Law of Sines: $\frac{\sin A}{a} = \frac{\sin C}{c}$ $\frac{\sin A}{8} = \frac{\sin 60°}{7}$ $\sin A = \frac{8 \sin 60°}{7} = \frac{8 \times \frac{\sqrt{3}}{2}}{7} = \frac{4\sqrt{3}}{7} \approx 0.9897$ $A = \arcsin(0.9897) \approx 81.8°$

Verification:

• $c^2 = 49$ gives $c = 7$ ✓
• When $C = 60°$, $\cos 60° = 0.5$ ✓

$14^2 = 8^2 + 11^2 - 2(8)(11)\cos C$

$196 = 64 + 121 - 176\cos C$

$196 = 185 - 176\cos C$

$11 = -176\cos C$

$\cos C = -\frac{11}{176} = -0.0625$

$C = \cos^{-1}(-0.0625) \approx 93.6°$

Part (b): Area formula using two sides and included angle:

First, we need to find one of the other angles. Let's find angle $A$:

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

$64 = 121 + 196 - 2(11)(14)\cos A$

$64 = 317 - 308\cos A$

$-253 = -308\cos A$

$\cos A = \frac{253}{308} = 0.821$

$A = \cos^{-1}(0.821) \approx 34.7°$

Now use: Area $= \frac{1}{2}bc\sin A$

$= \frac{1}{2}(11)(14)\sin 34.7°$

$= \frac{1}{2}(154)(0.569)$

$\approx 43.8$ square units

Step 1: Identify the case

This is SSS (all three sides), so use the Law of Cosines.

Step 2: Find angle $C$ (largest angle, opposite longest side)

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

Substitute: $\cos C = \frac{20^2 + 15^2 - 25^2}{2(20)(15)}$ $\cos C = \frac{400 + 225 - 625}{600}$ $\cos C = \frac{0}{600} = 0$

Therefore: $C = \arccos(0) = 90°$

This is a right triangle!

Step 3: Find angle $A$

Now we can use Law of Sines: $\frac{\sin A}{a} = \frac{\sin C}{c}$ $\frac{\sin A}{20} = \frac{\sin 90°}{25}$ $\frac{\sin A}{20} = \frac{1}{25}$ $\sin A = \frac{20}{25} = \frac{4}{5} = 0.8$ $A = \arcsin(0.8) \approx 53.1°$

Step 4: Find angle $B$

$B = 180° - A - C = 180° - 53.1° - 90° = 36.9°$

Answer:

• $A \approx 53.1°$
• $B \approx 36.9°$
• $C = 90°$

Verification:

• $A + B + C \approx 53.1° + 36.9° + 90° = 180°$ ✓
• Check Pythagorean theorem: $20^2 + 15^2 = 400 + 225 = 625 = 25^2$ ✓
• This confirms it's a right triangle! ✓

Note: This is a 3-4-5 right triangle scaled by 5 (sides 15, 20, 25).