Problem 18 - Algebra 2 and Trigonometry

Question

A field is bordered by two pairs of parallel roads so that the shape of the field is a parallelogram. The lengths of two adjacent sides of the field are 2 kilometers and 3 kilometers, and the length of the shorter diagonal of the field is 3 kilometers. a. Find the cosine of the acute angle of the parallelogram. b. Find the exact value of the sine of the acute angle of the parallelogram. c. Find the exact value of the area of the field. d. Find the area of the field to the nearest integer.

Step-by-Step Solution

Answer

a) \( \cos(\theta) = \frac{1}{3} \); b) \( \sin(\theta) = \frac{2\sqrt{2}}{3} \); c) \( 4\sqrt{2} \text{ km}^2 \); d) 6 km².

Step 1: Understanding the Parallelogram

We have a parallelogram with side lengths 2 km and 3 km, and the shorter diagonal measures 3 km. Let's denote the sides as \( AB = 2 \) km, \( BC = 3 \) km, and the diagonal \( AC = 3 \) km.

Step 2: Applying the Cosine Rule to Find Cosine

Using the cosine rule for triangle \( ABC \), \( AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(\theta) \). Substituting the known values: \( 3^2 = 2^2 + 3^2 - 2 \times 2 \times 3 \times \cos(\theta) \). Simplifying gives \( 9 = 4 + 9 - 12 \cos(\theta) \). Thus, \( 12 \cos(\theta) = 4 \), so \( \cos(\theta) = \frac{1}{3} \).

Step 3: Finding the Sine Using the Identity

To find \( \sin(\theta) \), use the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \). Given \( \cos(\theta) = \frac{1}{3} \), we have \( \sin^2(\theta) = 1 - \left(\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9} \). Therefore, \( \sin(\theta) = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \).

Step 4: Calculating the Exact Area of the Parallelogram

The area \( A \) of the parallelogram can be calculated using the formula \( A = ab \sin(\theta) \), where \( a = 2 \) km, \( b = 3 \) km, and \( \sin(\theta) = \frac{2\sqrt{2}}{3} \). Hence, \( A = 2 \times 3 \times \frac{2\sqrt{2}}{3} = 4\sqrt{2} \text{ square kilometers} \).

Step 5: Rounding the Area to the Nearest Integer

Calculate \( 4\sqrt{2} \) using \( \sqrt{2} \approx 1.414 \). Therefore, the approximate area is \( 4 \times 1.414 = 5.656 \text{ square kilometers} \), which rounds to \( 6 \text{ square kilometers} \).

Key Concepts

Cosine Rule

The cosine rule is a fundamental principle in trigonometry used for solving triangles, especially non-right angled ones. It generalizes the Pythagorean theorem to all triangles and is particularly useful in calculating an angle or a side when other sides are known. For any triangle with sides \(a\), \(b\), and \(c\) opposite to angles \(\alpha\), \(\beta\), and \(\gamma\) respectively, the formula is:

• \(c^2 = a^2 + b^2 - 2ab\cos(\gamma)\)
• \(b^2 = a^2 + c^2 - 2ac\cos(\beta)\)
• \(a^2 = b^2 + c^2 - 2bc\cos(\alpha)\)

In the context of our parallelogram problem, this rule helps us find the angle between two sides when we know the lengths of both sides and the length of a diagonal. By rearranging the formula of the cosine rule, we can solve for the cosine of the angle as part of the solution.

Sine and Cosine Identities

Sine and cosine identities are key components in trigonometry that relate the angles and sides of triangles. One essential identity that relates the sine and cosine of any angle \(\theta\) is:

• \(\sin^2(\theta) + \cos^2(\theta) = 1\)

This identity is crucial in our calculations because once we determined \(\cos(\theta)\) using the cosine rule, we can directly substitute it into the identity to find \(\sin(\theta)\). In our exercise, after calculating \(\cos(\theta) = \frac{1}{3}\), we use the identity to find \(\sin^2(\theta)\), which allows us to determine \(\sin(\theta)\) as \(\frac{2\sqrt{2}}{3}\). These identities provide a powerful framework for solving various trigonometric problems.

Area of a Parallelogram

The area of a parallelogram can be determined using the formula involving the base, height, and the sine of the angle between them. When the lengths of the sides and the sine of the included angle are known, the formula is:

• \(A = a \times b \times \sin(\theta)\)

Here, \(a\) and \(b\) are the lengths of the sides, and \(\theta\) is the angle between them. In the given problem, using the calculated \(\sin(\theta)\), we plug in \(a = 2\) km and \(b = 3\) km to get the exact area as \(4\sqrt{2}\) square kilometers. This formula shows how different components of geometry and trigonometry come together to solve real-world problems involving areas.

Problem-Solving in Geometry

Problem-solving in geometry often involves using various mathematical concepts and rules to derive solutions. An effective problem-solving strategy is to break down the problem as we did by understanding the shape and its properties first, in this case, the parallelogram. For geometry, one must:

• Understand the given information and what needs to be found.
• Determine which geometric rules and trigonometric identities are applicable.
• Use logical steps to apply these rules and identities.
• Double-check calculations and use estimation for verification.

In our exercise, we started by identifying the key components of the parallelogram and applied the cosine rule and trigonometric identities to find the area, demonstrating the integration of different geometric properties in solving complex problems.