We are given the following data for a triangle:

• $c = 60$
• $\angle A = 41^\circ$
• $\angle B = 25^\circ$

We need to solve the triangle using the Law of Sines. The Law of Sines states:

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Step 1: Find angle $\angle C$

The sum of the angles in any triangle is $180^\circ$. Therefore:

$\angle C = 180^\circ - \angle A - \angle B = 180^\circ - 41^\circ - 25^\circ = 114^\circ$

Step 2: Apply the Law of Sines

Now that we know all three angles, we can use the Law of Sines to find the sides $a$ and $b$.

First, express the Law of Sines for side $a$ and side $b$:

$\frac{a}{\sin A} = \frac{c}{\sin C} \quad \text{and} \quad \frac{b}{\sin B} = \frac{c}{\sin C}$

We can now substitute the known values:

$\frac{a}{\sin 41^\circ} = \frac{60}{\sin 114^\circ} \quad \text{and} \quad \frac{b}{\sin 25^\circ} = \frac{60}{\sin 114^\circ}$

Step 3: Solve for $a$ and $b$

Let’s compute the values.

Solve for $a$:

$a = \frac{60 \cdot \sin 41^\circ}{\sin 114^\circ}$

Solve for $b$:

$b = \frac{60 \cdot \sin 25^\circ}{\sin 114^\circ}$

Now I'll compute the values.The side lengths of the triangle, rounded to two decimal places, are:

• $a \approx 43.09$
• $b \approx 27.76$

Would you like further details or have any questions?

Here are some related questions for practice:

1. How would you apply the Law of Cosines to solve a triangle?
2. What if $\angle C$ was obtuse? How does that affect the solution?
3. Can you use the Law of Sines when two angles and no sides are given?
4. How does the Law of Sines differ from the Law of Cosines?
5. How would you approach solving a right triangle using the Law of Sines?

Tip: Always check if the sum of the angles equals 180° after solving for the third angle to avoid calculation errors.