Problem 39 - A Graphical Approach to Precalculus with Limits

Question

Determine the number of triangles ABC possible with the given parts. \(a=50, b=61, A=58^{\circ}\)

Step-by-Step Solution

Answer

No triangle is possible since \( \sin(B) > 1 \).

Step 1: Use the Law of Sines

Use the Law of Sines to find the possible values for angle B. According to the Law of Sines, we have: \[ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} \] Substitute the known values: \[ \frac{50}{\sin(58^{\circ})} = \frac{61}{\sin(B)} \] Simplify and solve for \( \sin(B) \):\[ \sin(B) = \frac{61 \cdot \sin(58^{\circ})}{50} \] Calculate \( \sin(58^{\circ}) \) using a calculator: \( \sin(58^{\circ}) \approx 0.848 \)Then compute:\[ \sin(B) = \frac{61 \cdot 0.848}{50} \approx 1.035 \]

Step 2: Analyze the Solution

The value of \( \sin(B) = 1.035 \) is greater than 1, which is not possible since the sine of an angle cannot exceed 1. This means that no triangle ABC can be formed with these given parts.

Key Concepts

Understanding the Law of Sines

The Law of Sines is a fundamental concept in trigonometry, often used to solve triangles in cases where side lengths and angles are involved. This law establishes a relationship between the lengths of sides of a triangle and the sines of its angles. In our specific example, the Law of Sines is stated as follows:

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

This formula is particularly useful in cases such as the given exercise, where two sides and an angle opposite one of the sides are known. It allows for the calculation of other unknown angles or sides.
In the exercise, we begin by setting up the proportion using the given side lengths and angle. By substituting the values for side \( a \) and angle \( A \), along with side \( b \), we can solve for \( \sin(B) \). This involves using a calculator to find the sine of angle \( A \), then calculating \( \sin(B) \). If \( \sin(B) \) exceeds 1, it indicates an error or impossibility, as no angle can have a sine value greater than 1.

Key Triangle Properties

Triangles are one of the basic shapes in geometry, and they have some important properties that can help when solving problems like the one given in the exercise. One of these key properties involves the angles of a triangle - the sum of the angles in any triangle is always \( 180^{\circ} \).
In addition, the side lengths are also tied closely with angles. Opposite sides and angles have a particular relationship, as described by the Law of Sines. An impossible value, like \( \sin(B) \) being greater than 1, suggests that the given sides and angle cannot form a real triangle.

• The side opposite the larger angle is the longest, and vice versa.
• The shortest side is opposite the smallest angle.

These properties guide us in understanding the relationships in a triangle and how to check for the possibility of a triangle's existence.

Calculating Angles with Confidence

Calculating angles in triangles can sometimes seem daunting, especially involving trigonometric functions. First, ensure you've correctly understood the problem by listing the given sides and angles.
In our example, after calculating \( \sin(B) \), we determine if it is a valid angle. Since trigonometric functions like sine are bounded (restricted from \(-1\) to \(1\)), any value calculated outside this range suggests a mistake or impossibility.
If \( \sin(B) \) were within this range, we would proceed by using the inverse sine function to find \( B \),

• \[ B = \sin^{-1}(\sin(B)) \]

This value would be checked against the properties of triangles to ensure sum constraints and side relationships are followed. However, when \( \sin(B) > 1 \), as in this problem, it confirms that a triangle cannot exist with the given dimensions. Such clarity ensures confidence in determining whether forming a triangle is feasible.