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 smaller than ∠B2.)
a = 76,    b = 100,    ∠A = 23°
∠B1 =
°
∠B2 =
°
∠C1 =
°
∠C2 =
°
c1 =

c2 =

To solve this problem using the Law of Sines, we need to use the given values:

- \( a = 76 \)
- \( b = 100 \)
- \( \angle A = 23^\circ \)

### Step 1: Use the Law of Sines to find \( \angle B_1 \)

The Law of Sines states:
\[
\frac{a}{\sin(\angle A)} = \frac{b}{\sin(\angle B)}
\]
Substituting the known values:
\[
\frac{76}{\sin(23^\circ)} = \frac{100}{\sin(\angle B_1)}
\]
Solving for \( \sin(\angle B_1) \):
\[
\sin(\angle B_1) = \frac{100 \times \sin(23^\circ)}{76}
\]
We will now calculate this value.

### Step 2: Determine the possible values for \( \angle B_1 \) and \( \angle B_2 \)

Once we have \( \sin(\angle B_1) \), we'll determine whether there are two possible values for \( \angle B \) (i.e., \( \angle B_1 \) and \( \angle B_2 \)) using the sine function's property that \( \sin(\theta) = \sin(180^\circ - \theta) \).

Let me calculate this now.We have calculated the following values:

- \( \sin(\angle B_1) \approx 0.514 \)
- \( \angle B_1 \approx 30.9^\circ \)
- \( \angle B_2 = 180^\circ - 30.9^\circ \approx 149.1^\circ \)

### Step 3: Calculate \( \angle C_1 \) and \( \angle C_2 \)

Using the fact that the sum of the angles in a triangle is \( 180^\circ \), we can find \( \angle C_1 \) and \( \angle C_2 \):
\[
\angle C_1 = 180^\circ - \angle A - \angle B_1
\]
\[
\angle C_2 = 180^\circ - \angle A - \angle B_2
\]

Let me calculate these now.The remaining angles are:

- \( \angle C_1 \approx 126.1^\circ \)
- \( \angle C_2 \approx 7.9^\circ \)

### Step 4: Use the Law of Sines to find \( c_1 \) and \( c_2 \)

We will use the Law of Sines again to find the side \( c \):
\[
\frac{a}{\sin(\angle A)} = \frac{c_1}{\sin(\angle C_1)}
\]
\[
\frac{a}{\sin(\angle A)} = \frac{c_2}{\sin(\angle C_2)}
\]

Now, I will calculate \( c_1 \) and \( c_2 \).The values for the sides are:

- \( c_1 \approx 157.2 \)
- \( c_2 \approx 26.9 \)

### Final Answer:

- \( \angle B_1 \approx 30.9^\circ \)
- \( \angle B_2 \approx 149.1^\circ \)
- \( \angle C_1 \approx 126.1^\circ \)
- \( \angle C_2 \approx 7.9^\circ \)
- \( c_1 \approx 157.2 \)
- \( c_2 \approx 26.9 \)

If you have any questions or need further clarification, feel free to ask!

Here are 5 questions to expand on this:

1. How does the Law of Cosines relate to solving triangles when no angles are given?
2. Can a triangle exist with \( \angle B_1 = 90^\circ \) given the sides provided?
3. What happens if \( \angle A \) were larger, say \( 60^\circ \)? How would that affect the solution?
4. Why is it possible to have two solutions (with \( B_1 \) and \( B_2 \)) in this case?
5. What are the differences between the Law of Sines and the Law of Tangents for solving triangles?

**Tip:** The Law of Sines works well when dealing with non-right triangles where angles and side lengths are partially known.

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 smaller than ∠B2.) a = 76, b = 100, ∠A = 23° ∠B1 = ° ∠B2 = ° ∠C1 = ° ∠C2 = ° c1 = c2 =

Learn how to use the Law of Sines to solve a triangle with given sides a = 76, b = 100, and angle ∠A = 23°. This problem explains how to calculate multiple triangle solutions, including ∠B1, ∠B2, ∠C1, ∠C2, c1, and c2. Suitable for students in Grades 10-12.

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