Problem 23
Question
Determine whether each triangle has no solution, one solution, or two solutions. Then solve each triangle. Round measures of sides to the nearest tenth and measures of angles to the nearest degree. $$ A=124^{\circ}, a=1, b=2 $$
Step-by-Step Solution
Verified Answer
The triangle has no solution because the calculated \( \sin B \) is not possible.
1Step 1: Check the Validity of the Triangle Using the Law of Sines
The Law of Sines states that \( \frac{a}{\sin A} = \frac{b}{\sin B} \). Substituting the known values, we have \( \frac{1}{\sin 124^{\circ}} = \frac{2}{\sin B} \). First, calculate \( \sin 124^{\circ} \approx 0.829\). Hence, \( \frac{1}{0.829} \approx 1.206 \). For the triangle to be valid, \( \frac{2}{\sin B} \approx 1.206 \) must be satisfied. This gives \( \sin B = \frac{2}{1.206} \approx 1.659 \), which is not possible because the sine of an angle cannot exceed 1.
2Step 2: Conclusion about the Triangle
Since \( \sin B \) cannot exceed 1, and we found \( \sin B \approx 1.659 \), this triangle is invalid. Thus, it has no solutions. There are no real possible angle measures for B that satisfy the requirements of the triangle, so no solution exists for such a configuration.
Key Concepts
Solving TrianglesSine FunctionTriangle Validity
Solving Triangles
Understanding how to solve triangles is crucial in geometry. It's essentially about figuring out all the unknown parts of a triangle, given some initial information like side lengths or angles. When it comes to solving triangles, we often use trigonometric laws, such as the Law of Sines or the Law of Cosines, especially for non-right triangles.
In the given problem, we start with two sides and a non-included angle, which is known as the SSA (Side-Side-Angle) condition. This specific configuration is notoriously troublesome because it can lead to different possibilities:
- No solution
- One unique solution
- Two possible solutions
Sine Function
The sine function is a foundational concept in trigonometry. It relates the ratio of the length of the side opposite to an angle to the length of the hypotenuse in a right triangle. For any angle θ, it is represented as \( ext{sin} heta \).In this specific problem, the sine function helps determine the possible values of the unknown angles in a triangle. Using the Law of Sines, \( rac{a}{ ext{sin} A} = rac{b}{ ext{sin} B} \), we expect \( ext{sin} B \amequation = rac{2}{1.206} \) for the triangle to be viable.However, sine values always lie between -1 and 1. If calculations yield a sine value like 1.659, it's physically impossible, since it exceeds the maximum allowable value. This mathematical outcome already indicates issues with forming a plausible triangle using the given measurements.
Triangle Validity
The concept of triangle validity is about confirming whether a given set of conditions can construct a triangle. Triangles have specific criteria for being valid, grounded in geometry and trigonometry.One essential rule is that no angle's sine value should go beyond 1, as theoretically and practically it's unachievable. In the context of the given problem, when substituting the known values into the Law of Sines, \( ext{sin} B = rac{2}{1.206} \approx 1.659 \), it explicitly indicates a violation of the rules, making a valid triangle impossible.Hence, before solving a triangle, validating the initial input values to ensure compliance with mathematical limits is vital. This thorough check avoids wasted effort in attempting to solve an impossible scenario and confirms that the given conditions are ready for further calculations.
Other exercises in this chapter
Problem 22
Find the exact value of each trigonometric function. \(\sin 240^{\circ}\)
View solution Problem 22
Rewrite each degree measure in radians and each radian measure in degrees. \(-15^{\circ}\)
View solution Problem 23
Find each value. Write angle measures in radians. Round to the nearest hundredth. $$ \sin \left(\operatorname{Arctan} \frac{\sqrt{3}}{3}\right) $$
View solution Problem 23
Determine whether each triangle should be solved by beginning with the Law of Sines or Law of Cosines. Then solve each triangle. Round measures of sides to the
View solution