Problem 13
Question
Solve for the remaining side(s) and angle(s), if possible, using any appropriate technique. $$ a=16, \alpha=63^{\circ}, b=20 $$
Step-by-Step Solution
Verified Answer
A triangle with the given dimensions doesn't exist because \( \sin \beta > 1 \), indicating a calculation beyond possible real values.
1Step 1: Identify the Given Information
We are given one side and the angles opposite two of the sides in a triangle: \( a = 16 \), \( \alpha = 63^{\circ} \), and \( b = 20 \). We need to find side \( c \) and angle \( \beta \).
2Step 2: Calculate Angle β using the Law of Sines
Start by using the Law of Sines: \( \frac{a}{\sin \alpha} = \frac{b}{\sin \beta} \). Substitute known values: \( \frac{16}{\sin 63^{\circ}} = \frac{20}{\sin \beta} \). Calculate \( \sin \beta \) as \( \sin \beta = \frac{20 \times \sin 63^{\circ}}{16} \).
3Step 3: Solve for Angle β
Calculate \( \sin 63^{\circ} \approx 0.8910 \), so \( \sin \beta = \frac{20 \times 0.8910}{16} \approx 1.1138 \). However, \( \sin \beta \) cannot be greater than 1, which indicates a problem since this would mean there's no possible triangle with the given dimensions.
4Step 4: Analyze the Result
The result \( \sin \beta > 1 \) implies that no such triangle can exist with the sides \( a = 16 \), \( b = 20 \), and the angle \( \alpha = 63^{\circ} \). This is the ambiguous case of the SSA condition, where the given dimensions do not form a valid triangle.
Key Concepts
Triangle SSA ConditionTrigonometric FunctionsTriangle Solving Techniques
Triangle SSA Condition
The Triangle SSA (Side-Side-Angle) condition creates an intriguing scenario in geometry. It arises when we know two sides and an angle that is not included between them. This situation is often called the "ambiguous case" because it might yield zero, one, or two possible triangles. In our given problem, we have side length \( a = 16 \), angle \( \alpha = 63^{\circ} \), and side \( b = 20 \). Here, we note the angle \( \alpha \) is known, but the angle \( \beta \) opposite side \( b \) needs to be determined.SSA can be tricky because of the potential for these differing outcomes. Key factors to consider include the angle size (whether it is acute or obtuse), and the relation between the side opposite the known angle (\( a \)) and the other given side (\( b \)). This condition demands careful analysis using trigonometric principles.
Trigonometric Functions
Trigonometric functions are pivotal in solving triangles, especially under the SSA condition. The Law of Sines is a tool employed to find unknown angles or sides in a triangle by setting a proportion between sides and their opposite angles:\[\frac{a}{\sin(\alpha)} = \frac{b}{\sin(\beta)} = \frac{c}{\sin(\gamma)}\]In this problem, we used the Law of Sines to establish a relationship between \( a \), \( \alpha \), \( b \), and \( \beta \). By substituting the known values:\[\frac{16}{\sin(63^{\circ})} = \frac{20}{\sin(\beta)}\]However, as we found \( \sin(\beta) \) greater than 1 for the calculated ratio, this suggests an issue. Since the range of the sine function is from -1 to 1, any calculation where \( \sin(\beta) > 1 \) is impossible, indicating a non-existent triangle. Trigonometric functions, therefore, provide crucial checks in verifying the feasibility of a triangular setup.
Triangle Solving Techniques
Solving triangles, especially under the SSA condition, requires strategic application of trigonometric laws and careful interpretation of results. Several techniques can be deployed:
- Use the Law of Sines: Establish relationships between sides and corresponding opposite angles.
- Check Trigonometric Limits: Ensure that the values fall within the acceptable range of trigonometric functions (e.g., \( \sin \beta \leq 1 \)).
- Beyond the Basics: If initial checks fail (like in our scenario where \( \sin \beta > 1 \)), the SSA condition must be re-evaluated for possible configurations or adjustments.
Other exercises in this chapter
Problem 13
Find a polar representation for the complex number \(z\) and then identify \(\operatorname{Re}(z)\), \(\operatorname{Im}(z),|z|, \arg (z)\) and \(\operatorname{
View solution Problem 13
Graph the following equations. $$ r=\frac{4}{1+3 \cos (\theta)} $$
View solution Problem 13
Plot the point given in polar coordinates and then give three different expressions for the point such that (a) \(r0\) and \(\theta \leq 0\) (c) \(r>0\) and \(\
View solution Problem 14
In Exercises \(1-20\), plot the set of parametric equations by hand. Be sure to indicate the orientation imparted on the curve by the parametrization. $$ \left\
View solution