Problem 22 - Algebra and Trigonometry

Question

\(19-28\) . Use the Law of Sines to solve for all possible triangles that satisfy the given conditions. $$ b=45, \quad c=42, \quad \angle C=38^{\circ} $$

Step-by-Step Solution

Answer

Two triangles are possible: \((A, B, a) = (100.58^\circ, 41.42^\circ, 67.12)\) and \((A, B, a) = (3.42^\circ, 138.58^\circ, 4.08)\).

Step 1: Identify Given Information

We are given two sides of the triangle, \(b = 45\) and \(c = 42\), and an angle \(\angle C = 38^\circ\). We need to determine the other angles and the remaining side, \(a\).

Step 2: Apply the Law of Sines

Using the Law of Sines, the equation is \( \frac{b}{\sin B} = \frac{c}{\sin C} = \frac{a}{\sin A} \). Starting with \( \frac{c}{\sin C} \), we have: \( \frac{42}{\sin 38^\circ} \).

Step 3: Calculate \( \sin B \)

Now, we need \( \sin B = \frac{b}{\frac{c}{\sin C}} = \frac{45}{\frac{42}{\sin 38^\circ}}\). This simplifies to \( \sin B = \frac{45 \cdot \sin 38^\circ}{42}\).

Step 4: Solve for Angle \( B \)

Calculate \( \sin B \) and find the inverse sine. \( \sin B \approx \frac{45 \cdot 0.6157}{42} = 0.6607 \). Thus, \( B \approx \sin^{-1}(0.6607) \approx 41.42^\circ\).

Step 5: Check for the Second Possible Triangle

Since the sine function is positive in both the first and second quadrants, check if \( B = 180^\circ - 41.42^\circ = 138.58^\circ \) forms a valid triangle (\(A + B + C < 180^\circ\)).

Step 6: Calculate Angle \( A \) for Both Cases

For the first case, \( A = 180^\circ - B - C = 180^\circ - 41.42^\circ - 38^\circ = 100.58^\circ\). For the second case, \( A = 180^\circ - 138.58^\circ - 38^\circ = 3.42^\circ\).

Step 7: Calculate Side \( a \) Using the Law of Sines

For both scenarios, use the Law of Sines. First case: \( a = \frac{c \cdot \sin A}{\sin C} = \frac{42 \cdot \sin 100.58^\circ}{\sin 38^\circ} \). Second case: \( a = \frac{42 \cdot \sin 3.42^\circ}{\sin 38^\circ} \).

Step 8: Calculate the Length of \( a \)

For the first triangle case, \( a \approx \frac{42 \cdot 0.9848}{0.6157} = 67.12 \). For the second triangle, due to very small \( \sin A \), \( a \approx \frac{42 \cdot 0.0599}{0.6157} = 4.08 \).

Step 9: Conclusion on Possible Triangles

Both triangles are possible. The first triangle has \( A = 100.58\), \( B = 41.42\), \( C = 38\), \( a = 67.12\). The second triangle has \( A = 3.42\), \( B = 138.58\), \( C = 38\), \( a = 4.08\).

Key Concepts

Triangle Solving

When you're faced with solving a triangle, like in the given exercise, the main goal is to find all the missing sides and angles when some parts are known. For this task, we use the Law of Sines because it is especially helpful when we know:

• Two sides and a non-included angle (SSA scenario)
• Two angles and a non-included side (AAS or ASA scenarios)

In the exercise, we know two sides, 45 and 42, and an angle, 38°, opposite one of the sides. By establishing relationships through the Law of Sines, which states \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \), where a, b, and c are sides, and A, B, and C are their respective opposite angles, we can systematically determine all unknowns.
Calculating each missing angle and side carefully helps ensure no possible triangles are overlooked. Each triangle's side lengths and angles are checked for mathematical and logical consistency to ensure they form a valid triangle. This method is essential for solving triangle problems accurately.

Ambiguous Case

The ambiguous case occurs when using the Law of Sines with two sides and a non-included angle, also known as the SSA condition. Here, the possibility arises for having more than one valid triangle.
In this scenario:

• If given \( b = 45 \), \( c = 42 \), and \( \angle C = 38^{\circ} \), the angle opposite side "b" or "c" might produce two different angles "B" due to the nature of the sine function.
• The sine function is positive in both the first and second quadrants, leading to two potential angles for each sine value.

This means one needs to calculate:

• One angle using direct sine inverse
• The alternative angle by subtracting from 180°

Checking if both results can create valid triangles involves ensuring their internal angles sum to less than 180°, which confirms the existence of two triangles.

Trigonometric Identities

Trigonometric identities lay the foundation for solving triangles through functions like sine, cosine, and tangent. These identities facilitate calculations by providing necessary conversions and checking results for solving angle and side values in triangles.
The Law of Sines utilizes the sine identity, notably for the ambiguous case. In this exercise, we apply this specific identity:

• \( \sin B = \frac{b \cdot \sin 38^{\circ}}{42} \) to find angle B.

Further, basic trigonometric principles ensure calculations are correct, straight through the solution process. For instance:

• Verifying angle sums
• Using reciprocal identities if needed

These identities ensure alliance with established mathematical rules, especially when checking calculated angles and sides for logical consistency in triangle formation.

Angle Calculation

Calculating angles is a crucial step in solving triangles. After identifying known angles and sides, determining the remaining angles involves applying trigonometric relationships.
In the exercise, once \( \sin B \) was found (around 0.6607), calculating angle \( B \) was straightforward using the inverse sine function: \( B \approx \sin^{-1}(0.6607) \approx 41.42^{\circ} \).
We also checked for another potential angle, \( 180^{\circ} - 41.42^{\circ} = 138.58^{\circ} \), as angles from the sine identity can manifest in two possible values in positive quadrants.

• For the first angle set, calculate angle \( A \) by subtracting sum of angles B and C from 180°: \( A = 180^{\circ} - B - C \).
• Repeat for any additional angles verified by the ambiguous case.

The calculated angles ensure the triangle's sides are legitimate, conserving the properties of triangular geometry.