Chapter 10: Solution of Triangles

Overview

The solution of triangles is a fundamental topic in trigonometry that involves finding unknown sides and angles of triangles using trigonometric relationships. This chapter explores three essential tools: the Sine Rule, Cosine Rule, and various area formulas. Mastery of these techniques is crucial for solving problems in geometry, physics, engineering, and navigation.

Learning Objectives

After completing this chapter, you will be able to:

• Apply the Sine Rule to solve triangles
• Use the Cosine Rule for various triangle configurations
• Calculate areas using trigonometric formulas
• Handle ambiguous cases in triangle solutions
• Apply these concepts to real-world problems

Key Concepts

10.1 Sine Rule

Statement of Sine Rule

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

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

Where R is the radius of the circumscribed circle.

Applications of Sine Rule

The Sine Rule is particularly useful when:

1. Two angles and one side are known (AAS or ASA cases)
2. Two sides and one angle are known (SSA case)

Ambiguous Case (SSA)

When given two sides and a non-included angle, there can be:

• No solution: If a < b sin A
• One solution: If a = b sin A or a ≥ b
• Two solutions: If b sin A < a < b

10.2 Cosine Rule

Statement of Cosine Rule

For any triangle ABC:

1. For side a: $a^2 = b^2 + c^2 - 2bc \cos A$
2. For side b: $b^2 = a^2 + c^2 - 2ac \cos B$
3. For side c: $c^2 = a^2 + b^2 - 2ab \cos C$

Applications of Cosine Rule

The Cosine Rule is particularly useful when:

1. Three sides are known (SSS case)
2. Two sides and the included angle are known (SAS case)

Rearranged Forms for Angles

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

10.3 Area of a Triangle

Standard Area Formula

$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$

Trigonometric Area Formulas

1. Two sides and included angle:

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

2. Three sides (Heron's Formula):

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

Where s = (a + b + c)/2 is the semi-perimeter.

3. Using coordinates (Shoelace Formula):

$$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

Visual Learning

Important Formulas and Methods

Key Triangle Solution Formulas

Problem-Solving Strategies

Triangle Solution Steps:

1. Identify known information (sides/angles)
2. Choose appropriate rule based on given info
3. Apply formula(s) systematically
4. Check for multiple solutions in SSA case
5. Verify triangle properties (angle sum = 180°)

Area Calculation Steps:

1. Identify available information
2. Choose appropriate area formula
3. Calculate step by step
4. Include units in final answer

Solved Examples

Example 1: Sine Rule (AAS Case)

In triangle ABC, A = 45°, B = 60°, and a = 8 cm. Find the remaining sides and angle C.

Solution:

First, find angle C: C = 180° - A - B = 180° - 45° - 60° = 75°

Now apply Sine Rule:

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

Find b:

$$\frac{8}{\sin 45°} = \frac{b}{\sin 60°}$$ $$b = \frac{8 \times \sin 60°}{\sin 45°} = \frac{8 \times \sqrt{3}/2}{\sqrt{2}/2} = \frac{8\sqrt{3}}{\sqrt{2}} = \frac{8\sqrt{6}}{2} = 4\sqrt{6} \text{ cm}$$

Find c:

$$\frac{8}{\sin 45°} = \frac{c}{\sin 75°}$$ $$c = \frac{8 \times \sin 75°}{\sin 45°}$$

sin 75° = sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30° = $(\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) + (\frac{\sqrt{2}}{2})(\frac{1}{2}) = \frac{\sqrt{2}}{2} (\frac{\sqrt{3}}{2} + \frac{1}{2}) = \frac{\sqrt{2}}{2} \times \frac{\sqrt{3} + 1}{2} = \frac{\sqrt{2}(\sqrt{3} + 1)}{4}$

$$c = \frac{8 \times \sqrt{2}(\sqrt{3} + 1)/4}{\sqrt{2}/2} = \frac{2\sqrt{2}(\sqrt{3} + 1)}{\sqrt{2}/2} = 2\sqrt{2}(\sqrt{3} + 1) \times \frac{2}{\sqrt{2}} = 4(\sqrt{3} + 1) \text{ cm}$$

Example 2: Sine Rule (Ambiguous Case)

In triangle ABC, a = 10 cm, b = 12 cm, and A = 30°. Find all possible solutions.

Solution:

Check for ambiguous case: b sin A = 12 × sin 30° = 12 × 0.5 = 6 cm

Since 6 < 10 < 12 (b sin A < a < b), there are two possible solutions.

First solution: Using Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B}$

$$\frac{10}{\sin 30°} = \frac{12}{\sin B}$$ $$\frac{10}{0.5} = \frac{12}{\sin B}$$ $$20 = \frac{12}{\sin B} \Rightarrow \sin B = \frac{12}{20} = 0.6 \Rightarrow B \approx 36.87°$$

C = 180° - 30° - 36.87° = 113.13° c = $\frac{a \times \sin C}{\sin A} = \frac{10 \times \sin 113.13°}{0.5} \approx \frac{10 \times 0.9239}{0.5} \approx 18.48 \text{ cm}$

Second solution: B = 180° - 36.87° = 143.13° C = 180° - 30° - 143.13° = 6.87° c = $\frac{10 \times \sin 6.87°}{0.5} \approx \frac{10 \times 0.1197}{0.5} \approx 2.39 \text{ cm}$

Example 3: Cosine Rule (SAS Case)

In triangle ABC, b = 5 cm, c = 7 cm, and A = 60°. Find side a and the area.

Solution:

Use Cosine Rule for side a:

$$a^2 = b^2 + c^2 - 2bc \cos A$$ $$a^2 = 5^2 + 7^2 - 2(5)(7) \cos 60°$$ $$a^2 = 25 + 49 - 70(0.5)$$ $$a^2 = 74 - 35 = 39$$ $$a = \sqrt{39} \approx 6.24 \text{ cm}$$

Use area formula with two sides and included angle:

$$\text{Area} = \frac{1}{2}bc \sin A = \frac{1}{2}(5)(7) \sin 60° = \frac{1}{2} \times 35 \times \frac{\sqrt{3}}{2} = \frac{35\sqrt{3}}{4} \approx 15.15 \text{ cm}^2$$

Example 4: Cosine Rule (SSS Case)

In triangle ABC, a = 8 cm, b = 7 cm, c = 5 cm. Find all angles.

Solution:

Use Cosine Rule for each angle:

Angle A:

$$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{7^2 + 5^2 - 8^2}{2 \times 7 \times 5} = \frac{49 + 25 - 64}{70} = \frac{10}{70} = \frac{1}{7} \approx 0.1429$$ $$A = \cos^{-1}(\frac{1}{7}) \approx 81.79°$$

Angle B:

$$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{8^2 + 5^2 - 7^2}{2 \times 8 \times 5} = \frac{64 + 25 - 49}{80} = \frac{40}{80} = 0.5$$ $$B = \cos^{-1}(0.5) = 60°$$

Angle C:

$$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{8^2 + 7^2 - 5^2}{2 \times 8 \times 7} = \frac{64 + 49 - 25}{112} = \frac{88}{112} = \frac{11}{14} \approx 0.7857$$ $$C = \cos^{-1}(\frac{11}{14}) \approx 38.21°$$

Check: 81.79° + 60° + 38.21° = 180° ✓

Example 5: Heron's Formula

Find the area of a triangle with sides 6 cm, 8 cm, and 10 cm.

Solution:

Use Heron's formula: s = (6 + 8 + 10)/2 = 24/2 = 12 cm

$$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{12(12-6)(12-8)(12-10)} = \sqrt{12 \times 6 \times 4 \times 2} = \sqrt{576} = 24 \text{ cm}^2$$

Note: This is a right triangle (6-8-10 is 3-4-5 scaled), so area = $\frac{1}{2} \times 6 \times 8 = 24 \text{ cm}^2$ ✓

Mathematical Derivations

Derivation of Sine Rule

Consider triangle ABC with altitude h from C to AB.

In right triangle ADC: h = b sin A In right triangle BDC: h = a sin B

Therefore: b sin A = a sin B ⇒ $\frac{a}{\sin A} = \frac{b}{\sin B}$

Similarly, by drawing other altitudes, we get $\frac{a}{\sin A} = \frac{c}{\sin C}$

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

Derivation of Cosine Rule

Consider triangle ABC with coordinates: A(0,0), B(c,0), C(b cos A, b sin A)

Distance BC = a:

$$a^2 = (b \cos A - c)^2 + (b \sin A - 0)^2$$ $$a^2 = b^2 \cos^2 A - 2bc \cos A + c^2 + b^2 \sin^2 A$$ $$a^2 = b^2(\cos^2 A + \sin^2 A) + c^2 - 2bc \cos A$$ $$a^2 = b^2 + c^2 - 2bc \cos A$$

Derivation of Heron's Formula

Start with area = $\frac{1}{2}bc \sin A$ Using $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$, we get $\sin A = \sqrt{1 - \cos^2 A}$ After algebraic manipulation and simplification, we obtain:

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

Real-World Applications

1. Surveying and Land Measurement

Surveyors use:

• Triangulation: Measuring distances to inaccessible points
• Area calculations: Determining land area for ownership
• Height measurements: Calculating building heights

Example: To measure distance across a river, place stakes A and B on one bank, and measure angles from stakes C and D on the same bank to stake E on the opposite bank.

2. Navigation and Positioning

Navigation applications:

• Bearing calculations: Determining directions
• Distance estimation: Calculating travel distances
• Position fixing: Locating vessels or aircraft

Example: A ship travels 20 km on bearing 045°, then 30 km on bearing 120°. Use triangle methods to find final position relative to start.

3. Architecture and Construction

Building applications:

• Structural design: Calculating angles and forces
• Roof pitch: Determining optimal angles
• Foundation layout: Precise positioning of structures

4. Physics and Engineering

Physics applications:

• Force resolution: Breaking forces into components
• Projectile motion: Calculating trajectories
• Mechanical advantage: Analyzing simple machines

Complex Problem-Solving Techniques

Problem: Find the area of triangle ABC if AB = 10 cm, AC = 8 cm, and the angle between AB and the median to AC is 45°.

Solution:

Let M be the midpoint of AC, so AM = MC = 4 cm.

In triangle ABM, we have:

• AB = 10 cm
• AM = 4 cm
• Angle between AB and AM = 45°

Using the area formula for triangle ABM:

$$\text{Area} = \frac{1}{2} \times AB \times AM \times \sin(\text{angle between them})$$ $$= \frac{1}{2} \times 10 \times 4 \times \sin 45° = 20 \times \frac{\sqrt{2}}{2} = 10\sqrt{2} \text{ cm}^2$$

Since M is midpoint, triangles ABM and CBM have equal areas. Total area of ABC = 2 × 10√2 = 20√2 c$m^2$

Problem: In triangle ABC, find the maximum possible area if perimeter is fixed at P = 30 cm.

Solution:

For a given perimeter, the equilateral triangle has maximum area.

If ABC is equilateral: a = b = c = 30/3 = 10 cm

Using Heron's formula: s = 30/2 = 15 cm

$$\text{Area} = \sqrt{15(15-10)(15-10)(15-10)} = \sqrt{15 \times 5 \times 5 \times 5} = \sqrt{1875} = 25\sqrt{3} \approx 43.30 \text{ cm}^2$$

This is the maximum possible area for perimeter 30 cm.

Problem: Find the angle between the diagonals of a rectangle with sides 6 cm and 8 cm.

Solution:

In rectangle ABCD with AB = 6 cm, BC = 8 cm: Diagonals AC = BD = $\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ cm

Consider triangle AOB where O is intersection of diagonals: AO = BO = 5 cm (half of diagonal)

Using Cosine Rule:

$$\cos \theta = \frac{AO^2 + BO^2 - AB^2}{2 \times AO \times BO} = \frac{25 + 25 - 36}{2 \times 5 \times 5} = \frac{50 - 36}{50} = \frac{14}{50} = \frac{7}{25} = 0.28$$ $$\theta = \cos^{-1}(0.28) \approx 73.74°$$

Therefore, angle between diagonals is approximately 73.74°.

Summary Points

• Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ (best for AAS, ASA, SSA cases)
• Cosine Rule: $a^2 = b^2 + c^2 - 2bc \cos A$ (best for SAS, SSS cases)
• Ambiguous case: SSA can have 0, 1, or 2 solutions
• Area formulas: Multiple options depending on given information
• Real applications: Surveying, navigation, architecture, physics

Common Mistakes to Avoid

1. Rule selection errors - Choose Sine vs Cosine based on given information
2. Ambiguous case oversight - Always check for multiple solutions in SSA case
3. Angle sum errors - Remember triangle angles sum to 180°
4. Unit confusion - Keep consistent units for sides and angles
5. Calculation errors - Use calculator carefully, especially for inverse trig functions

SPM Exam Tips

Exam Strategies

1. Memorize formulas - Know both Sine and Cosine rules accurately
2. Identify given information - Choose the right rule based on what's given
3. Handle ambiguous cases - Check conditions for multiple solutions
4. Show working clearly - Step-by-step solutions for partial marks
5. Check triangle validity - Ensure angles sum to 180°

Key Exam Topics

• Sine Rule applications (30% of questions)
• Cosine Rule applications (30% of questions)
• Area calculations (25% of questions)
• Ambiguous cases (10% of questions)
• Practical applications (5% of questions)

Time Management Tips

• Sine Rule problems: 5-6 minutes
• Cosine Rule problems: 6-7 minutes
• Area calculations: 4-5 minutes
• Complex applications: 8-10 minutes

Practice Problems

Level 1: Sine Rule

1. In triangle ABC, A = 30°, B = 45°, a = 8 cm. Find b and c.

2. In triangle PQR, P = 60°, Q = 75°, q = 12 cm. Find p and r.

Level 2: Cosine Rule

3. In triangle XYZ, x = 5 cm, y = 7 cm, Z = 60°. Find z.

4. In triangle ABC, a = 8 cm, b = 10 cm, c = 12 cm. Find all angles.

Level 3: Area Calculations

5. Find the area of triangle with sides 5 cm, 6 cm, and 7 cm using Heron's formula.

6. Triangle has sides 8 cm and 10 cm with included angle 60°. Find area.

7. Triangle has vertices A(1,2), B(4,6), C(7,2). Find area using coordinate method.

Level 4: Ambiguous Cases

8. In triangle ABC, a = 10 cm, b = 12 cm, A = 30°. Find all possible solutions.

9. In triangle PQR, p = 15 cm, q = 20 cm, P = 40°. Determine if ambiguous and find solutions.

Level 5: Applications

10. Surveying: From point A, two trees B and C are observed. Angle BAC = 45°, AB = 100 m, AC = 120 m. Find distance BC.

11. Navigation: A ship sails 30 km on bearing 030°, then 40 km on bearing 120°. Find distance from starting point.

12. Physics: Forces of 50 N and 70 N act at an angle of 60°. Find resultant force and its direction.

Did You Know?

The Sine Rule and Cosine Rule were known to ancient civilizations. The Sine Rule was used by Greek astronomers as early as the 3rd century BCE, while the Cosine Rule was developed by Islamic mathematicians in the Middle Ages. These formulas revolutionized navigation, astronomy, and surveying, enabling precise calculations that were previously impossible.

Quick Reference Guide

---

Mastering the solution of triangles provides essential tools for geometry, physics, and real-world problem-solving. These trigonometric techniques will serve as foundation for advanced topics in calculus and vector analysis.