Examples of Solving Problems Using the Law of Sines and Cosines

Introduction

The law of sines and the law of cosines are fundamental principles in trigonometry that help solve various geometric problems, particularly in non-right triangles. These laws allow you to find unknown sides and angles, making them invaluable in fields such as engineering, architecture, and navigation. Below are three diverse examples that demonstrate how to apply these laws effectively.

Example 1: Finding a Side Length in a Triangle

Context

In a situation where you need to determine the length of a side in a triangle, the law of sines can be extremely useful. This is common in fields like construction, where accurate measurements are crucial.

Given a triangle ABC where:

• Angle A = 30°
• Angle B = 45°
• Side a (opposite angle A) = 10 meters

To find side b (opposite angle B), we can use the law of sines:

Calculation

Using the law of sines:

\[ \frac{a}{\sin A} = \frac{b}{\sin B} \]

Substituting the known values:

\[ \frac{10}{\sin(30°)} = \frac{b}{\sin(45°)} \]

Calculating:

• \( \sin(30°) = 0.5 \)
• \( \sin(45°) = \frac{\sqrt{2}}{2} \approx 0.7071 \)

Now substituting:

\[ \frac{10}{0.5} = \frac{b}{0.7071} \]

This simplifies to:

\[ 20 = \frac{b}{0.7071} \]

Now, multiplying both sides by 0.7071:

\[ b = 20 * 0.7071 \approx 14.14 ext{ meters} \]

Notes

This method is particularly useful for finding unknown sides when two angles and one side are known. Variations of this problem can involve different angle configurations and side lengths, but the approach remains the same.

Example 2: Calculating an Angle in a Triangle

Context

In navigation or surveying, you often need to determine unknown angles in a triangle formed by landmarks or survey points. The law of cosines is particularly useful for this purpose.

Consider triangle ABC, with:

• Side a = 7 meters
• Side b = 10 meters
• Side c = 5 meters

We want to find angle C.

Calculation

Using the law of cosines:

\[ c^2 = a^2 + b^2 - 2ab \cdot \cos C \]

Substituting the known values:

\[ 5^2 = 7^2 + 10^2 - 2(7)(10) \cdot \cos C \]

Calculating:

• \( 25 = 49 + 100 - 140 \cdot \cos C \)
• \( 25 = 149 - 140 \cdot \cos C \)
• Rearranging gives:
• \( 140 \cdot \cos C = 149 - 25 \)
• \( 140 \cdot \cos C = 124 \)

Now, dividing by 140:

\[ \cos C = \frac{124}{140} \approx 0.8857 \]

To find angle C:

\[ C = \cos^{-1}(0.8857) \approx 27.15° \]

Notes

This application of the law of cosines is particularly useful when all three sides of a triangle are known. Variations may involve different combinations of sides and angles, but the formula remains a reliable tool.

Example 3: Area Calculation of a Triangle Using the Law of Sines

Context

In fields like architecture or landscaping, calculating the area of a triangle when you only have two sides and the included angle is a common task. The law of sines can be adapted to help with this.

Consider a triangle ABC with:

• Side a = 8 meters
• Side b = 6 meters
• Included angle C = 60°

Calculation

The area (A) of the triangle can be calculated using the formula:

\[ A = \frac{1}{2}ab \sin C \]

Substituting the known values:

\[ A = \frac{1}{2}(8)(6) \sin(60°) \]

Calculating:

• \( \sin(60°) = \frac{\sqrt{3}}{2} \approx 0.8660 \)

So:

\[ A = \frac{1}{2}(48)(0.8660) \approx 20.784 \]

Thus, the area of triangle ABC is approximately 20.78 square meters.

Notes

This method provides a quick way to find the area of a triangle when two sides and the included angle are known. Variations could include different angles or side lengths, but the area formula remains applicable.

These examples illustrate practical applications of the law of sines and cosines, demonstrating their importance in solving geometric problems effectively.