Lesson 7.2: The Sine and Cosine Rules

Introduction

In our previous lessons, we learned about right-angled triangles and basic trigonometric functions. In this lesson, we will expand our knowledge by introducing the sine and cosine rules, which are essential for solving non-right-angled triangles. Understanding these rules will not only help you in mathematics but also in fields like physics, engineering, and architecture where triangle calculations are crucial.

Learning Objectives

By the end of this lesson, students will be able to:

1. Understand and apply the sine rule for non-right-angled triangles.
2. Understand and apply the cosine rule for non-right-angled triangles.
3. Calculate the area of a triangle using the formula $A = \frac{1}{2}ab \sin C$.
4. Choose the appropriate rule for solving a given triangle problem.
5. Solve for unknown sides and angles in non-right-angled triangles using either the sine or cosine rule.

The Sine Rule

The sine rule relates the lengths of sides of a triangle to the sines of its angles. It states that in any triangle:

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

Where:

• $a$, $b$, and $c$ are the lengths of the sides opposite to angles $A$, $B$, and $C$, respectively.

Understanding the Sine Rule

To develop an intuition for the sine rule, let's consider a triangle $ABC$. The ratio of the length of a side to the sine of the angle opposite that side is constant for all three sides.

For example, this is particularly useful when we know:

• Two angles and one side (AAS or ASA)
• Two sides and a non-included angle (SSA)

Worked Example of the Sine Rule

Example Problem: In triangle $ABC$, if $A = 30^\circ$, $B = 60^\circ$, and $a = 10$, find the length of side $b$.

1. Step 1: Use the sine rule formula:

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

1. Step 2: Plug in the known values:

$$ \frac{10}{\sin 30^\circ} = \frac{b}{\sin 60^\circ} $$

1. Step 3: Calculate the sine values:

$$ \sin 30^\circ = \frac{1}{2} \quad \text{and} \quad \sin 60^\circ = \frac{\sqrt{3}}{2} $$

1. Step 4: Substitute the sine values into the equation:

$$ \frac{10}{\frac{1}{2}} = \frac{b}{\frac{\sqrt{3}}{2}} $$

1. Step 5: Simplify the equation:

$$ 20 = \frac{b}{\frac{\sqrt{3}}{2}} \Rightarrow b = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} $$

So, the length of side $b$ is $10\sqrt{3}$.

The Cosine Rule

The cosine rule relates the lengths of the sides of a triangle to the cosine of one of its angles. It is particularly useful for solving triangles when:

• All three sides are known (SSS)
• Two sides and the included angle are known (SAS)

It states that in any triangle:

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

Understanding the Cosine Rule

When you know two sides and the angle in between them, you can calculate the length of the third side. Conversely, if you know all three sides, you can find an angle.

Worked Example of the Cosine Rule

Example Problem: In triangle $ABC$, let $a = 7$, $b = 5$, and $C = 60^\circ$. Find the length of side $c$.

1. Step 1: Use the cosine rule formula:

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

1. Step 2: Plug in the known values:

$$ c^2 = 7^2 + 5^2 - 2(7)(5) \cos 60^\circ $$

1. Step 3: Calculate the cosine value:

$$ \cos 60^\circ = \frac{1}{2} $$

1. Step 4: Substitute into the equation:

$$ c^2 = 49 + 25 - 2(7)(5) \times \frac{1}{2} $$

1. Step 5: Simplify the equation:

$$ c^2 = 49 + 25 - 35 = 39 $$

1. Step 6: Solve for $c$:

$$ c = \sqrt{39} $$

Thus, the length of side $c$ is $\sqrt{39}$.

Area of a Triangle Using Trigonometry

The area $A$ of triangle $ABC$ can also be calculated when two sides and the included angle are known, using the formula:

$$ A = \frac{1}{2}ab \sin C $$

Worked Example of Area Calculation

Example Problem: Given that $a = 8$, $b = 6$, and $C = 45^\circ$, find the area of triangle $ABC$.

1. Step 1: Use the area formula:

$$ A = \frac{1}{2}ab \sin C $$

1. Step 2: Plug in the known values:

$$ A = \frac{1}{2}(8)(6) \sin 45^\circ $$

1. Step 3: Calculate the sine value:

$$ \sin 45^\circ = \frac{\sqrt{2}}{2} $$

1. Step 4: Substitute this into the formula:

$$ A = \frac{1}{2}(8)(6) \times \frac{\sqrt{2}}{2} $$

1. Step 5: Simplify the area calculation:

$$ A = \frac{1}{2}(48) \times \frac{\sqrt{2}}{2} = 24 \times \frac{\sqrt{2}}{2} = 12\sqrt{2} $$

So, the area of triangle $ABC$ is $12\sqrt{2}$ square units.

Choosing the Appropriate Rule

Choosing the sine or cosine rule often depends on the information given in a problem:

• Use the sine rule when you have:
• Two angles and one side (AAS or ASA)
• Two sides and a non-included angle (SSA)
• Use the cosine rule when you have:
• Two sides and the included angle (SAS)
• All three sides (SSS)

Conclusion

In this lesson, we explored the sine and cosine rules and their applications. We learned to solve for unknown sides and angles in non-right-angled triangles, and we calculated the area using trigonometry. Mastery of these concepts will greatly enhance your ability to tackle a wide range of problems in mathematics and its applications.

Study Notes

• Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
• Cosine Rule: $c^2 = a^2 + b^2 - 2ab \cos C$
• Area Formula: $A = \frac{1}{2}ab \sin C$
• Choosing Rules:
• Sine rule for AAS, ASA, SSA problems.
• Cosine rule for SAS, SSS problems.