Two sides and an angle are given below. Determine whether the given information results in one​ triangle, two​ triangles, or no triangle at all. Solve any resulting​ triangle(s).
b equals 8 commab=8, c equals 9 commac=9, Upper B equals 30 degreesB=30°
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Part 1
Select the correct choice below​ and, if​ necessary, fill in the answer boxes to complete your choice.
​(Type an integer or decimal rounded to two decimal places as​ needed.)
A.
A single triangle is​ produced, where Calmost equals≈enter your response heredegrees°​, Aalmost equals≈enter your response heredegrees°​, and aalmost equals≈enter your response here.
B.
Two triangles are​ produced, where the triangle with the smaller angle C has Upper C 1C1almost equals≈enter your response heredegrees°​, Upper A 1A1almost equals≈enter your response heredegrees°​, and a 1a1almost equals≈enter your response here​, and the triangle with the larger angle C has Upper C 2C2almost equals≈enter your response heredegrees°​, Upper A 2A2almost equals≈enter your response heredegrees°​, and a 2a2almost equals≈enter your response here.
C.
No triangles are produced.

To determine whether the given information results in one triangle, two triangles, or no triangle at all, we need to apply the Law of Sines to solve for the unknown angles and sides. Given are:

- \( b = 8 \)
- \( c = 9 \)
- \( B = 30^\circ \)

### Step 1: Use the Law of Sines to solve for angle \( C \)
\[
\frac{b}{\sin(B)} = \frac{c}{\sin(C)}
\]
Substitute the known values:
\[
\frac{8}{\sin(30^\circ)} = \frac{9}{\sin(C)}
\]
Since \( \sin(30^\circ) = 0.5 \), this simplifies to:
\[
\frac{8}{0.5} = \frac{9}{\sin(C)}
\]
\[
16 = \frac{9}{\sin(C)}
\]
Now solve for \( \sin(C) \):
\[
\sin(C) = \frac{9}{16} = 0.5625
\]

### Step 2: Check for the number of possible triangles
Since \( \sin(C) = 0.5625 \), this gives us two possible solutions for angle \( C \):
- \( C_1 = \sin^{-1}(0.5625) \approx 34.22^\circ \)
- \( C_2 = 180^\circ - 34.22^\circ \approx 145.78^\circ \)

#### Case 1: \( C_1 = 34.22^\circ \)
Now, to find angle \( A \), use the fact that the sum of the angles in a triangle is 180°:
\[
A_1 = 180^\circ - B - C_1 = 180^\circ - 30^\circ - 34.22^\circ = 115.78^\circ
\]
This is a valid triangle since all angles are positive and less than 180°.

#### Case 2: \( C_2 = 145.78^\circ \)
For the second triangle:
\[
A_2 = 180^\circ - B - C_2 = 180^\circ - 30^\circ - 145.78^\circ = 4.22^\circ
\]
This is also a valid triangle, though \( A_2 \) is quite small.

### Step 3: Solve for side \( a \)
Using the Law of Sines again to solve for side \( a \) in both cases:

For the first triangle:
\[
\frac{a_1}{\sin(A_1)} = \frac{b}{\sin(B)}
\]
\[
\frac{a_1}{\sin(115.78^\circ)} = \frac{8}{\sin(30^\circ)}
\]
\[
\frac{a_1}{0.9063} = \frac{8}{0.5}
\]
\[
a_1 = \frac{8 \times 0.9063}{0.5} \approx 14.50
\]

For the second triangle:
\[
\frac{a_2}{\sin(A_2)} = \frac{b}{\sin(B)}
\]
\[
\frac{a_2}{\sin(4.22^\circ)} = \frac{8}{0.5}
\]
\[
\frac{a_2}{0.0736} = 16
\]
\[
a_2 = 16 \times 0.0736 \approx 1.18
\]

### Conclusion:
There are **two triangles** formed:
- For the first triangle:
  - \( C_1 \approx 34.22^\circ \)
  - \( A_1 \approx 115.78^\circ \)
  - \( a_1 \approx 14.50 \)

- For the second triangle:
  - \( C_2 \approx 145.78^\circ \)
  - \( A_2 \approx 4.22^\circ \)
  - \( a_2 \approx 1.18 \)

Would you like further explanation on any part of the process?

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**Related Questions:**
1. What is the Law of Sines, and how is it applied to non-right triangles?
2. Can there be situations where no triangle is formed when given two sides and an angle?
3. How do you handle ambiguous cases in trigonometry?
4. What is the difference between SSA, SAS, and ASA triangle configurations?
5. How do you find the area of a triangle when given two sides and an angle?

**Tip:** The Law of Sines is particularly useful for solving triangles when two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA) are known.

Two sides and an angle are given below. Determine whether the given information results in one triangle, two triangles, or no triangle at all. Solve any resulting triangle(s). b=8, c=9, B=30°. Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

Learn how to determine whether the given two sides and an angle result in one, two, or no triangles using the Law of Sines. This detailed solution covers triangle properties and trigonometric formulas, solving for angles and side lengths step by step. Suitable for high school students in Grades 10-12.

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