Use the Law of Sines to solve for all possible triangles that satisfy the given conditions. (If an answer does not exist, enter DNE. Round your answers to one decimal place. Below, enter your answers so that ∠*B*1 is larger than ∠*B*2.)

*a* = 36,    *c* = 48,    ∠*A* = 31°

∠*B*1 =

  °

    

∠*B*2 =

  °

∠*C*1 =

  °

    

∠*C*2 =

  °

*b*1 =

    

*b*2 =

To solve the triangle using the Law of Sines, we will start by calculating the possible values for angle \( B \) and then determine the other sides and angles of the triangle.

### Given:
- \( a = 36 \)
- \( c = 48 \)
- \( \angle A = 31^\circ \)

### Step 1: Use the Law of Sines
The Law of Sines states:
\[
\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}
\]
Substitute the known values for \( a \), \( c \), and \( \angle A \):
\[
\frac{36}{\sin(31^\circ)} = \frac{48}{\sin(C)}
\]
First, calculate the value of \( \sin(31^\circ) \):
\[
\sin(31^\circ) \approx 0.515
\]
Thus,
\[
\frac{36}{0.515} \approx 69.9
\]
Now, apply the Law of Sines to find \( \sin(C) \):
\[
\frac{69.9}{48} = \frac{1}{\sin(C)}
\]
This gives:
\[
\sin(C) = \frac{48}{69.9} \approx 0.686
\]

### Step 2: Solve for \( \angle C \)
Using the inverse sine function, calculate \( \angle C \):
\[
\angle C_1 = \sin^{-1}(0.686) \approx 43.4^\circ
\]
Now, calculate the other possible solution for \( \angle C \) (since sine has two possible values in the range of 0 to 180 degrees):
\[
\angle C_2 = 180^\circ - 43.4^\circ \approx 136.6^\circ
\]

### Step 3: Solve for \( \angle B \)
Using the fact that the sum of angles in a triangle is \( 180^\circ \), calculate \( \angle B_1 \) and \( \angle B_2 \):

For \( \angle B_1 \):
\[
\angle B_1 = 180^\circ - 31^\circ - 43.4^\circ = 105.6^\circ
\]

For \( \angle B_2 \):
\[
\angle B_2 = 180^\circ - 31^\circ - 136.6^\circ = 12.4^\circ
\]

### Step 4: Solve for \( b_1 \) and \( b_2 \)
Using the Law of Sines again, solve for \( b_1 \) and \( b_2 \):

For \( b_1 \):
\[
\frac{b_1}{\sin(105.6^\circ)} = \frac{36}{\sin(31^\circ)}
\]
\[
b_1 = \frac{36 \cdot \sin(105.6^\circ)}{0.515} \approx \frac{36 \cdot 0.965}{0.515} \approx 67.4
\]

For \( b_2 \):
\[
\frac{b_2}{\sin(12.4^\circ)} = \frac{36}{\sin(31^\circ)}
\]
\[
b_2 = \frac{36 \cdot \sin(12.4^\circ)}{0.515} \approx \frac{36 \cdot 0.215}{0.515} \approx 15.0
\]

### Final Answers:
- \( \angle B_1 = 105.6^\circ \)
- \( \angle B_2 = 12.4^\circ \)
- \( \angle C_1 = 43.4^\circ \)
- \( \angle C_2 = 136.6^\circ \)
- \( b_1 = 67.4 \)
- \( b_2 = 15.0 \)

Let me know if you'd like further details or have any questions.

Here are five related questions for further exploration:
1. How does the Law of Cosines differ from the Law of Sines?
2. What happens when there are no possible triangles?
3. How do we determine if a triangle is ambiguous using the Law of Sines?
4. Can the Law of Sines be used in right triangles?
5. How would the calculations change if \( \angle A \) was larger?

**Tip:** The Law of Sines can yield two possible triangles, but one must always check the validity of both using the angle sum property of triangles.

Use the Law of Sines to solve for all possible triangles that satisfy the given conditions. (If an answer does not exist, enter DNE. Round your answers to one decimal place. Below, enter your answers so that ∠B1 is larger than ∠B2.) a = 36, c = 48, ∠A = 31° ∠B1 = ° ∠B2 = ° ∠C1 = ° ∠C2 = ° b1 = b2 = 

Learn how to solve for all possible triangles using the Law of Sines with a = 36, c = 48, and ∠A = 31°. This trigonometry problem involves calculating angles B1, B2, C1, C2, and sides b1, b2, and understanding the ambiguous case of the Law of Sines. Suitable for students in Grades 10-12.

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