Problem 34 - Algebra 2

Question

Solve each triangle. Round measures of sides to the nearest tenth and measures of angles to the nearest degree. $$ B=18^{\circ}, C=142^{\circ}, b=20 $$

Step-by-Step Solution

Answer

A = 20°, a ≈ 22.1, c ≈ 57.2.

Step 1: Determine the Measure of Angle A

We know that the sum of angles in any triangle is \(180^\circ\). Therefore, to find the measure of angle \(A\), we can use the equation: \[A + B + C = 180^\circ\]. Substituting the known angles gives us: \[A + 18^\circ + 142^\circ = 180^\circ\]. Solving for \(A\), we have: \[A = 180^\circ - 18^\circ - 142^\circ = 20^\circ\]. This means \(A = 20^\circ\).

Step 2: Apply the Law of Sines

The Law of Sines is stated as: \[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\]. We'll first use this to find side \(a\). Using \(b = 20\), \(B = 18^\circ\), \(A = 20^\circ\), \(C = 142^\circ\), we set up the equation: \[\frac{a}{\sin 20^\circ} = \frac{20}{\sin 18^\circ}\]. Now, solving for \(a\): \[a = 20 \cdot \frac{\sin 20^\circ}{\sin 18^\circ}\]. Calculating this gives: \[a \approx 22.1\].

Step 3: Find the Remaining Side c

Next, we need to find side \(c\) using the Law of Sines again: \[\frac{c}{\sin 142^\circ} = \frac{20}{\sin 18^\circ}\]. Solving for \(c\), we have: \[c = 20 \cdot \frac{\sin 142^\circ}{\sin 18^\circ}\]. Calculating this gives: \[c \approx 57.2\].

Key Concepts

Law of Sines

The Law of Sines is a fundamental concept in trigonometry, especially useful for solving non-right triangles. It relates the lengths of the sides of a triangle to the sines of their opposite angles. This law is expressed as:

• \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \)

This means that the ratio of the length of a side to the sine of its opposite angle is the same for all three sides of the triangle. To apply this, you need to know at least one side length and its opposite angle. From there, you can find the remaining sides and angles easily, provided you have enough information about the other angles or sides.

In the given problem, we used the Law of Sines to solve for sides \(a\) and \(c\) after calculating angle \(A\). Knowing \(b = 20\), \(B = 18^\circ\), \(A = 20^\circ\), and \(C = 142^\circ\), allows us to set up proportionate equations for \(a\) and \(c\). This law is highly efficient for problems involving oblique triangles, where some angles and sides are known, but not all.

Angle Sum Property

The properties of triangles are fascinating, and one of the core tenets is the angle sum property. This property states that the sum of all interior angles in a triangle must always equal \(180^\circ\).

• \(A + B + C = 180^\circ\)

This principle is a cornerstone for solving triangles and helps verify that the triangle is valid. With two angles known, the third can be easily calculated by subtraction.

In this exercise, angle \(A\) was determined using this property. Starting with angles \(B = 18^\circ\) and \(C = 142^\circ\), we calculated angle \(A\) by subtracting the sum of \(B\) and \(C\) from \(180^\circ\). This straightforward calculation is crucial when beginning to solve any triangle, as it ensures all subsequent calculations are based on a proper triangle.

Solving Triangles

Becoming adept at solving triangles involves understanding the properties and relationships between angles and sides. This involves several strategies for finding unknown measures when given partial information.

• Use the angle sum property to find missing angles.
• Apply the Law of Sines to find unknown sides.
• Verify results through consistency with known triangle properties.

For the given problem, solving the triangle started by calculating the missing angle using the angle sum property. Once all angles were known, the Law of Sines was applied to determine the unknown side lengths. The process of solving a triangle is like piecing together a puzzle. Each known measure is a piece that helps reveal the unknown, and verifying each step ensures accuracy.

By understanding these fundamental principles, you will be able to solve any triangle, whether in a homework problem or in real-world applications.