Problem 7 - Trigonometry

Question

For each of the following triangles, solve for \(B\) and use the results to explain why the triangle has the given number of solutions. \(A=120^{\circ}, b=20 \mathrm{~cm}, a=30 \mathrm{~cm}\); one solution

Step-by-Step Solution

Answer

The triangle has one solution because \( B \approx 35.26^{\circ} \) fits with the angle constraints.

Step 1: Identify Triangle Components

We are given a triangle with angle \( A = 120^{\circ} \), side \( a = 30 \, \text{cm} \), and side \( b = 20 \, \text{cm} \). We need to find angle \( B \) and verify the number of solutions for the triangle.

Step 2: Use the Law of Sines

The Law of Sines states \( \frac{a}{\sin(A)} = \frac{b}{\sin(B)} \). Plugging in the given values: \[ \frac{30}{\sin(120^{\circ})} = \frac{20}{\sin(B)} \]. Calculate \( \sin(120^{\circ}) \) which is equal to \( \frac{\sqrt{3}}{2} \).

Step 3: Solve for \( \sin(B) \)

Rearrange the equation to find \( \sin(B) \):\[ \sin(B) = \frac{20 \cdot \sin(120^{\circ})}{30} = \frac{20 \cdot \frac{\sqrt{3}}{2}}{30} = \frac{20\sqrt{3}}{60} = \frac{\sqrt{3}}{3} \].

Step 4: Determine \( B \) from \( \sin(B) \)

Since \( \sin(B) = \frac{\sqrt{3}}{3} \), use a calculator or a trigonometric table to find that \( B \approx 35.26^{\circ} \).

Step 5: Verify Number of Solutions

In a triangle, the sum of angles is \(180^{\circ}\). Calculate the third angle \( C = 180^{\circ} - A - B \approx 180^{\circ} - 120^{\circ} - 35.26^{\circ} \approx 24.74^{\circ} \). Since both \( B \) and \( C \) are valid angles of the triangle and \( B \) found fits within this context, there is exactly one solution.

Key Concepts

Angle Sum Property

The angle sum property is a fundamental rule in geometry, especially for triangles. This principle states that the sum of the three interior angles of any triangle is always exactly 180 degrees. This rule is crucial when solving triangles, as it lets us calculate unknown angles once we have the measurements of the other angles.

In the given problem, we know that angle \( A = 120^{\circ} \) and angle \( B \) is approximately \( 35.26^{\circ} \). We can use the angle sum property to find the third angle \( C \) as follows:

• Sum the known angles \( A \) and \( B \): \( A + B = 120^{\circ} + 35.26^{\circ} = 155.26^{\circ} \).
• Subtract from 180 degrees: \( C = 180^{\circ} - 155.26^{\circ} = 24.74^{\circ} \).

So, the angle sum property confirms our calculations and further supports that there is exactly one valid solution for the triangle in this exercise.

Triangle Solutions

Determining the number of solutions for a triangle involves assessing whether the given measurements allow for one, two, or no possible triangles. In this exercise, we used the Law of Sines to find that \( \sin(B) = \frac{\sqrt{3}}{3} \), which corresponds to an angle \( B \approx 35.26^{\circ} \).

Here are key points to understand about triangle solutions:

• **One Solution**: This situation occurs when the given set of data results in a unique triangle, which is the case here because angle \( B \) and \( C \) add up harmoniously with \( A \) to make exactly 180 degrees.
• **Two Solutions**: Occur when two different triangles can be drawn from the given information, often happening when using the Law of Sines with an angle less than 90 degrees.
• **No Solution**: This arises when the given data conflicts with triangle properties, such as when the sides or angles do not comply with the triangle inequality theorem.

Understanding these conditions helps in systematically finding the correct number of solutions, reassuring students why this particular example only has one solution.

Trigonometric Identities

In solving for triangles using the Law of Sines, trigonometric identities often come into play. A trigonometric identity is an equality involving trigonometric functions that is true for every value of variables. They're critical in solving triangles as they help simplify expressions and solve equations.

In this exercise, we use one fundamental property:

• The sine of an angle: We calculated \( \sin(120^{\circ}) \) to be \( \frac{\sqrt{3}}{2} \). Knowing this identity allowed us to solve for \( \sin(B) \).

By breaking down trigonometric functions into known identities, it becomes easier to manipulate and solve for unknown values in triangle problems. Whether working with a calculator or a table of values, recognizing how to use trigonometric identities broadens the student's toolkit for addressing various geometry and trigonometry challenges.