Problem 23 - Algebra and Trigonometry

Question

\(19-28\) . Use the Law of Sines to solve for all possible triangles that satisfy the given conditions. $$ b=25, \quad c=30, \quad \angle B=25^{\circ} $$

Step-by-Step Solution

Answer

Two triangles: (1) Angles 25°, 30.53°, 124.47°; sides 54.14. (2) Angles 25°, 149.47°, 5.53°; sides 5.66.

Step 1: Understand the Problem

Given a triangle with sides \( b = 25 \), \( c = 30 \) and angle \( \angle B = 25^{\circ} \), we are to find all possible triangles using the Law of Sines. We will be solving for both angle \( A \) and \( C \), and side \( a \).

Step 2: Apply the Law of Sines

According to the Law of Sines, \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \). We know \( \frac{b}{\sin B} = \frac{25}{\sin 25^{\circ}} \).

Step 3: Find Angle C using Law of Sines

Using \( \sin C = \frac{c \cdot \sin B}{b} \), we have \( \sin C = \frac{30 \cdot \sin 25^{\circ}}{25} \approx 0.508 \). We calculate \( \angle C \) by finding \( \arcsin(0.508) \).

Step 4: Calculate Angle C

\( C \approx \arcsin(0.508) \approx 30.53^{\circ} \). Since \( \sin(150^{\circ} - C) = \sin C \), we check for the second possible angle: \( 149.47^{\circ} \).

Step 5: Determine Possible Triangles

For the first triangle, angles are \( \angle B = 25^{\circ} \), \( \angle C = 30.53^{\circ} \), and \( \angle A = 180^{\circ} - (25^{\circ} + 30.53^{\circ}) = 124.47^{\circ} \). For the second triangle (ambiguous case): \( \angle C' = 149.47^{\circ} \), \( \angle A' = 180^{\circ} - (25^{\circ} + 149.47^{\circ}) = 5.53^{\circ} \).

Step 6: Calculate Side a for Both Triangles

Using \( a = \frac{b \cdot \sin A}{\sin B} \), for the first triangle: \( a \approx \frac{25 \cdot \sin 124.47^{\circ}}{\sin 25^{\circ}} \approx 54.14 \). For the second triangle: \( a' \approx \frac{25 \cdot \sin 5.53^{\circ}}{\sin 25^{\circ}} \approx 5.66 \).

Step 7: Conclusion

Two possible triangles are found. The first triangle has angles \( 25^{\circ}, 30.53^{\circ}, 124.47^{\circ} \) and sides \( a \approx 54.14 \). The second triangle has angles \( 25^{\circ}, 149.47^{\circ}, 5.53^{\circ} \) and sides \( a' \approx 5.66 \).

Key Concepts

Triangle Properties

When solving any problem involving triangles, it's essential to familiarize yourself with basic triangle properties. Triangles have three sides and three angles. A core property is that the sum of all interior angles equals exactly 180 degrees. This property is crucial when calculating unknown angles because it provides a baseline for checking the validity of your calculations.

Triangles are categorized into various types based on side lengths, such as equilateral, isosceles, and scalene. Understanding these types helps in visualizing the triangle's structure when dealing with ambiguous cases or missing measurements. Another property is the law that governs exterior angles, stating that an exterior angle of a triangle is equal to the sum of the opposite interior angles. This can be useful in more complex problems or proofs.

In this specific exercise, we are focusing not only on calculating the unknown sides and angles but also considering the possibility of more than one triangle existing with the given dimensions. This is where understanding triangle properties helps prevent errors.

Angle Calculations

For angle calculations in triangles, especially when using the Law of Sines, it's helpful to understand the relationship between sine values and angles. The Law of Sines states that in any triangle, the ratio of each side length to the sine of its opposite angle is constant. Mathematically, it is expressed as: \[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\] Given side lengths and one angle, as in this exercise, we can find the other angles using this relationship. By rearranging this formula, you can solve for unknown angles using given side lengths and one known angle.

To calculate an unknown angle, say angle \(C\), we use the rearranged formula: \[\sin C = \frac{c \cdot \sin B}{b}\]This means you need the sine of a known angle (here, angle \(B\)), a known side (\(b\)), and a known opposite side (\(c\)). Once you find \(\sin C\), the angle \(C\) can be determined using the inverse sine function \(\arcsin\). The process helps in systematic and accurate angle calculation.

This exercise demonstrates detailed calculations that arise from utilizing the Law of Sines in practical scenarios, ensuring each derived angle is verified.

Ambiguous Case in Triangle

The ambiguous case arises in triangle problems when two different triangles could exist for the given measurements. This typically happens in SSA (Side-Side-Angle) conditions, like in our problem. In such cases, the calculated sine value could yield two possible angles, leading to different triangles.

To explore both possibilities, after finding one angle using the inverse sine function, you should also consider its supplementary angle. This is because sine of any angle equals sine of its supplementary angle (\[sin(180^{\circ} - \theta) = \sin \theta\]). For instance, if you find \(\angle C = 30.53^{\circ}\), there could be another valid triangle where \(\angle C = 149.47^{\circ}\), as long as all angles again sum to 180 degrees.

Once you have both potential values for an angle, compute the other angles and analyze any resulting triangles. Factors like possible negative angles or angle sums exceeding 180 degrees might lead to rejecting one triangle. In our working example considering SSA, both configurations led to valid triangles, highlighting the potential for multiple solutions in real-world geometry problems.