https://file.aiquickdraw.com/mathbot/file/1728763749867saf5fj3p.jpg

The image contains two geometry problems involving triangles. Let's analyze and solve both.

### Problem 1:
In triangle \( \Delta ABC \):
- \( b = 120 \)
- \( m(\angle A) = 32^\circ \)
- \( m(\angle C) = 48^\circ \)

You are asked to find sides \( AB \) and \( BC \).

### Solution to Problem 1:
1. Using the angle sum property of triangles:
   \[
   m(\angle B) = 180^\circ - (m(\angle A) + m(\angle C)) = 180^\circ - (32^\circ + 48^\circ) = 100^\circ
   \]
2. Use the Law of Sines to find the sides:
   \[
   \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
   \]
   Here, \( b = 120 \), so we calculate \( a \) and \( c \) as follows:
   - For side \( a \) (opposite angle \( \angle A \)):
     \[
     \frac{a}{\sin 32^\circ} = \frac{120}{\sin 100^\circ}
     \]
     Solve for \( a \):
     \[
     a = 120 \times \frac{\sin 32^\circ}{\sin 100^\circ}
     \]

   - For side \( c \) (opposite angle \( \angle C \)):
     \[
     \frac{c}{\sin 48^\circ} = \frac{120}{\sin 100^\circ}
     \]
     Solve for \( c \):
     \[
     c = 120 \times \frac{\sin 48^\circ}{\sin 100^\circ}
     \]

You can compute these values to get the exact lengths of sides \( AB \) and \( BC \).

---

### Problem 2:
In triangle \( \Delta ABC \):
- \( b = \sqrt{6} \) cm
- \( m(\angle A) = 75^\circ \)
- \( m(\angle C) = 45^\circ \)

You are asked to find the angles \( A \), \( B \), and \( C \), as well as the perimeter.

### Solution to Problem 2:
1. Use the angle sum property to find \( m(\angle B) \):
   \[
   m(\angle B) = 180^\circ - (m(\angle A) + m(\angle C)) = 180^\circ - (75^\circ + 45^\circ) = 60^\circ
   \]
2. Again, apply the Law of Sines to find sides:
   \[
   \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
   \]
   Since \( b = \sqrt{6} \), calculate \( a \) and \( c \).

   - For side \( a \) (opposite angle \( \angle A \)):
     \[
     \frac{a}{\sin 75^\circ} = \frac{\sqrt{6}}{\sin 60^\circ}
     \]
     Solve for \( a \):
     \[
     a = \sqrt{6} \times \frac{\sin 75^\circ}{\sin 60^\circ}
     \]
     
   - For side \( c \) (opposite angle \( \angle C \)):
     \[
     \frac{c}{\sin 45^\circ} = \frac{\sqrt{6}}{\sin 60^\circ}
     \]
     Solve for \( c \):
     \[
     c = \sqrt{6} \times \frac{\sin 45^\circ}{\sin 60^\circ}
     \]

Finally, calculate the perimeter by adding \( a + b + c \).

---

Would you like me to calculate the exact values for sides and perimeter? 

Here are some follow-up questions:
1. What are the other methods to solve triangles apart from the Law of Sines?
2. How does the Law of Cosines differ from the Law of Sines?
3. Can you calculate the angles of a triangle if two sides are known?
4. How can trigonometric identities help in solving non-right triangles?
5. What are the real-world applications of solving triangles using trigonometry?

**Tip:** Always check if the given triangle is valid using triangle inequality rules before solving!

Solve the following triangle problems:
1. In triangle ABC: b = 120, m(∠A) = 32°, and m(∠C) = 48°. Find sides AB and BC.
2. In triangle ABC: b = √6 cm, m(∠A) = 75°, and m(∠C) = 45°. Find angles A, B, and C, and then find the perimeter of the triangle.

Learn how to solve triangle problems using the Law of Sines and the Angle Sum Property. This guide covers calculating unknown sides and angles, with examples suitable for Grades 9-12. Problems include finding the sides of triangles and calculating perimeters using trigonometric principles.

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