An oblique triangle is any triangle that is not a right triangle (no 90° angle).

For oblique triangles, we cannot use basic trigonometry (SOH CAH TOA). Instead, we use:

• Law of Sines: Relates sides and their opposite angles
• Law of Cosines: Generalizes the Pythagorean theorem

Triangle Notation

For any triangle with vertices $A$, $B$, and $C$:

• Angles: $\angle A$, $\angle B$, $\angle C$ (or simply $A$, $B$, $C$)
• Sides:
  • Side $a$ is opposite angle $A$
  • Side $b$ is opposite angle $B$
  • Side $c$ is opposite angle $C$

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)

Step 1: Identify the case

This is AAS (two angles and one side), so use the Law of Sines.

Step 2: Find the third angle $C = 180° - A - B = 180° - 35° - 65° = 80°$

Step 3: Apply Law of Sines $\frac{a}{\sin A} = \frac{c}{\sin C}$

Substitute: $\frac{a}{\sin 35°} = \frac{10}{\sin 80°}$

Step 4: Solve for $a$ $a = \frac{10 \sin 35°}{\sin 80°}$

Calculate: $a = \frac{10 \times 0.5736}{0.9848} \approx \frac{5.736}{0.9848} \approx 5.82 \text{ cm}$

Answer: $a \approx 5.82$ cm

Verification:

• All angles sum to $180°$ ✓
• $a < c$ makes sense since $A < C$ (smaller angle opposite smaller side) ✓

a) Use the Law of Sines to find angle $B$. b) Find angle $C$. c) Find side $c$.

a) Use the Law of Cosines to find angle $C$. b) Find the area of the triangle.