Triangle ∆ABC, hac B = 40◦, b = 8cm, and c = 11cm.
Show that there are two possible values for C, and solve the triangle in each case. Solve the triangle means find all unknown sides and unknown angles.

Given the triangle \(\Delta ABC\) with the following information:

- \(\angle B = 40^\circ\)
- \(b = 8 \text{ cm}\)
- \(c = 11 \text{ cm}\)

We need to find the two possible values of \(\angle C\), and then solve the triangle for each case, i.e., find the unknown sides and angles.

### Step 1: Use the Law of Sines

We start with the Law of Sines:

\[
\frac{b}{\sin B} = \frac{c}{\sin C}
\]

Substitute the given values:

\[
\frac{8}{\sin 40^\circ} = \frac{11}{\sin C}
\]

### Step 2: Solve for \(\sin C\)

\[
\sin C = \frac{11 \times \sin 40^\circ}{8}
\]

Calculate \(\sin 40^\circ\) first:

\[
\sin 40^\circ \approx 0.6428
\]

Now, calculate \(\sin C\):

\[
\sin C = \frac{11 \times 0.6428}{8} \approx 0.8839
\]

### Step 3: Determine the Possible Values of \(\angle C\)

Since \(\sin C \approx 0.8839\), \(\angle C\) can have two possible values:

1. \(\angle C_1 = \arcsin(0.8839)\)
2. \(\angle C_2 = 180^\circ - \arcsin(0.8839)\)

Let's calculate these angles:

1. \(\angle C_1 \approx \arcsin(0.8839) \approx 62.3^\circ\)
2. \(\angle C_2 = 180^\circ - 62.3^\circ = 117.7^\circ\)

So, the two possible values for \(\angle C\) are \(62.3^\circ\) and \(117.7^\circ\).

### Step 4: Solve the Triangle for Each Case

#### Case 1: \(\angle C = 62.3^\circ\)

- Find \(\angle A\):
  \[
  \angle A = 180^\circ - \angle B - \angle C = 180^\circ - 40^\circ - 62.3^\circ = 77.7^\circ
  \]

- Use the Law of Sines to find \(a\):
  \[
  \frac{a}{\sin A} = \frac{b}{\sin B}
  \]
  \[
  a = \frac{8 \times \sin 77.7^\circ}{\sin 40^\circ}
  \]
  \[
  \sin 77.7^\circ \approx 0.9797
  \]
  \[
  a \approx \frac{8 \times 0.9797}{0.6428} \approx 12.2 \text{ cm}
  \]

So, the triangle with \(\angle C = 62.3^\circ\) has:

- \(\angle A = 77.7^\circ\)
- \(a \approx 12.2 \text{ cm}\)

#### Case 2: \(\angle C = 117.7^\circ\)

- Find \(\angle A\):
  \[
  \angle A = 180^\circ - \angle B - \angle C = 180^\circ - 40^\circ - 117.7^\circ = 22.3^\circ
  \]

- Use the Law of Sines to find \(a\):
  \[
  a = \frac{8 \times \sin 22.3^\circ}{\sin 40^\circ}
  \]
  \[
  \sin 22.3^\circ \approx 0.3790
  \]
  \[
  a \approx \frac{8 \times 0.3790}{0.6428} \approx 4.7 \text{ cm}
  \]

So, the triangle with \(\angle C = 117.7^\circ\) has:

- \(\angle A = 22.3^\circ\)
- \(a \approx 4.7 \text{ cm}\)

### Summary of Solutions

#### Case 1:
- \(\angle A = 77.7^\circ\)
- \(\angle B = 40^\circ\)
- \(\angle C = 62.3^\circ\)
- \(a \approx 12.2 \text{ cm}\)
- \(b = 8 \text{ cm}\)
- \(c = 11 \text{ cm}\)

#### Case 2:
- \(\angle A = 22.3^\circ\)
- \(\angle B = 40^\circ\)
- \(\angle C = 117.7^\circ\)
- \(a \approx 4.7 \text{ cm}\)
- \(b = 8 \text{ cm}\)
- \(c = 11 \text{ cm}\)

Would you like more details or have any other questions? Here are some related questions you might find interesting:

1. How do we know when there can be two possible solutions for a triangle?
2. Can a triangle have more than two possible solutions?
3. What happens if \(\sin C > 1\) when using the Law of Sines?
4. How do you apply the Law of Cosines in a similar problem?
5. How does the ambiguous case arise in the Law of Sines?

**Tip:** When solving triangles, always double-check if multiple solutions are possible, especially in cases where you use the Law of Sines.

Triangle ∆ABC, hac B = 40◦, b = 8cm, and c = 11cm. Show that there are two possible values for C, and solve the triangle in each case. Solve the triangle means find all unknown sides and unknown angles.

Explore the step-by-step solution of triangle ABC where angle B is 40°, side b is 8 cm, and side c is 11 cm. Learn how to find the two possible values for angle C using the Law of Sines and solve the triangle for each scenario. Suitable for high school students in Grades 10-12.

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