Let's solve this problem step by step using the Law of Sines.

Given:

• $m\angle B = 61^\circ$
• $AB = 4.28 \, \text{cm}$
• $AC = 3.87 \, \text{cm}$

Step 1: Use the Law of Sines to find $m\angle C$

The Law of Sines states: $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$ Where:

• $A, B, C$ are the angles of the triangle
• $a, b, c$ are the sides opposite to angles $A, B, C$ respectively

For this triangle: $\frac{\sin B}{AC} = \frac{\sin C}{AB}$

Substitute the known values: $\frac{\sin 61^\circ}{3.87} = \frac{\sin C}{4.28}$

Step 2: Solve for $\sin C$

$\sin C = \frac{4.28 \times \sin 61^\circ}{3.87}$ Let's calculate the value of $\sin C$.

$\sin 61^\circ \approx 0.8746$ $\sin C = \frac{4.28 \times 0.8746}{3.87} \approx 0.9669$

Step 3: Find possible values of $m\angle C$

Now, $\sin C = 0.9669$.

1. $m\angle C \approx \arcsin(0.9669) \approx 75.32^\circ$

However, since $\sin \theta = \sin (180^\circ - \theta)$, the second possible value for $m\angle C$ is: 2. $m\angle C = 180^\circ - 75.32^\circ = 104.68^\circ$

Thus, the two possible values for $m\angle C$ are: $m\angle C = 75.32^\circ, 104.68^\circ$

Step 4: Determine the appropriate value of $m\angle C$

To decide which value makes more sense, consider that the sum of angles in a triangle must equal $180^\circ$.

For the first case ($m\angle C = 75.32^\circ$): $m\angle A = 180^\circ - 61^\circ - 75.32^\circ = 43.68^\circ$

For the second case ($m\angle C = 104.68^\circ$): $m\angle A = 180^\circ - 61^\circ - 104.68^\circ = 14.32^\circ$

Assuming the diagram is to scale and based on typical geometric considerations, $m\angle A \approx 43.68^\circ$ seems more reasonable than $14.32^\circ$. So, the more appropriate value for $m\angle C$ is $75.32^\circ$.

Step 5: Find the length of $BC$

Using the Law of Sines again: $\frac{BC}{\sin A} = \frac{AC}{\sin C}$ Substitute the values for $AC$, $\sin C$, and $\sin A$: $BC = \frac{AC \times \sin A}{\sin C} = \frac{3.87 \times \sin 43.68^\circ}{\sin 75.32^\circ}$

Let's calculate this.

Final Answer

• $m\angle C = 75.32^\circ$
• $BC \approx 2.79 \, \text{cm}$

Would you like further details or any clarifications?

Here are 8 related questions:

1. What is the Law of Cosines, and when is it used?
2. How do you solve a triangle given two angles and one side (ASA case)?
3. What is the ambiguous case in the Law of Sines?
4. Can a triangle have two obtuse angles?
5. How does the Law of Sines handle obtuse triangles differently from acute triangles?
6. Why are there two possible angles for $m\angle C$ in this scenario?
7. How can the Law of Sines be applied to right triangles?
8. What are the limitations of the Law of Sines?

Tip: Always check for the possibility of two solutions when applying the Law of Sines in ambiguous cases.