Problem 17 - Precalculus

Question

The measures of two sides and an angle are given. Determine whether a triangle (or two) exist, and if so, solve the triangle(s). $$a=4, b=5, \alpha=16^{\circ}$$

Step-by-Step Solution

Answer

One triangle exists with \(\beta = 20.29^\circ\), \(\gamma = 143.71^\circ\), and \(c \approx 8.83\).

Step 1: Identify the given information

We are given a triangle with side lengths \(a = 4\) and \(b = 5\), and the angle \(\alpha = 16^\circ\) opposite side \(a\). This forms a situation referred to as the angle-side-side (ASS) ambiguity or the Law of Sines case.

Step 2: Calculate the angle opposite side b

Use the Law of Sines, which states \( \frac{a}{\sin(\alpha)} = \frac{b}{\sin(\beta)} \), to find \( \beta \). Rearrange to find \( \sin(\beta) = \frac{b \cdot \sin(\alpha)}{a} \). Substitute the given values: \( \sin(\beta) = \frac{5 \cdot \sin(16^\circ)}{4} \approx 0.3471 \).

Step 3: Determine the number of solutions for beta

Since \(\sin(\beta) \approx 0.3471\), \(\beta\) can equal \(\beta = \arcsin(0.3471) \approx 20.29^\circ\). Also, because sine is positive in the second quadrant, \(\beta\) could also equal \(180^\circ - 20.29^\circ = 159.71^\circ\).

Step 4: Check the triangle inequality

For the triangle inequalities, \(\alpha + \beta < 180^\circ\). Check for each potential \(\beta\): for \(\beta = 20.29^\circ\), \(16^\circ + 20.29^\circ < 180^\circ\), so this triangle is possible. For \(\beta = 159.71^\circ\), \(16^\circ + 159.71^\circ > 180^\circ\), so this triangle is not possible.

Step 5: Solve for the remaining angles and sides

With \(\beta = 20.29^\circ\), use the sum of angles in a triangle to find \(\gamma = 180^\circ - 16^\circ - 20.29^\circ = 143.71^\circ\). Use the Law of Sines again to find side \(c\): \(\frac{c}{\sin(\gamma)} = \frac{a}{\sin(\alpha)} \Rightarrow c = \frac{4 \cdot \sin(143.71^\circ)}{\sin(16^\circ)} \approx 8.83\).

Key Concepts

Angle-Side-Side (ASS) Ambiguity

The Angle-Side-Side (ASS) ambiguity arises when we are given a triangle with two sides and a non-included angle, like the information given in this exercise. This configuration is tricky because it doesn't always lead to a unique triangle.

Sometimes, you might have:

• No triangle at all.
• One possible triangle.
• Or even two possible triangles.

When you encounter this issue, the Law of Sines is your go-to tool for figuring out what triangles, if any, are possible. Simply put, the Law of Sines formula is used to compute the possible angles given the provided sides and one angle. For this, \[ \frac{a}{\sin(\alpha)} = \frac{b}{\sin(\beta)} \] has to be employed carefully, as even if it assists in finding another angle, verifying whether a triangle really can be built out of these measures is critical.

Triangle Inequality

The triangle inequality principle is a fundamental rule that determines whether three given line segments can form a triangle. According to this principle, the sum of any two sides of a triangle must be greater than the length of the third side.

This rule ensures the existence of a triangle with the given sides around the computed angles. For instance, if you're working on the possible triangle formed by the measures provided and you've calculated potential angles, the sum of the angles \[ \alpha + \beta + \gamma = 180^\circ \] must hold true for the triangle to exist. This relates back to checking whether the calculated angles indeed form a valid triangle, as seen in the solution steps when \[ \alpha + \beta < 180^\circ \] is ensured for possible triangles.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are fundamental to working with triangles, especially when involving angles and their relationships to the sides of the triangle. For example, in the context of this problem:

1. **Sine Function:** It helps determine unknown angles or sides by relating the sides opposite the angles. The function \[ \sin(\theta) \] provides a ratio involving the opposite side over the hypotenuse of a triangle.
2. **Inverse Sine:** Also known as arcsin, it is used to find angles when their sine value is known. Hence, when we found \[ \sin(\beta) \approx 0.3471 \] in the exercise, by using inverse sine, we found possible angles.

These trigonometric tools provide the necessary means to unlock the complexities and ambiguities often presented in problems involving triangles.