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.