https://file.aiquickdraw.com/mathbot/file/1729563181275ez2vrqfr.jpg

Let's break down and solve the three triangles one by one based on the given information.

### Triangle A:
- Given:
  - Angle \( A = 52^\circ \)
  - Side \( b = 70 \, \text{mm} \)
  - Side \( c = 84 \, \text{mm} \)
  
This is a non-right triangle, so we can use the **Law of Cosines** to find the third side, and the **Law of Sines** to find the remaining angles.

1. **Law of Cosines**:
   \[
   a^2 = b^2 + c^2 - 2bc \cdot \cos(A)
   \]
   Substituting the known values:
   \[
   a^2 = 70^2 + 84^2 - 2(70)(84) \cdot \cos(52^\circ)
   \]
   \[
   a^2 = 4900 + 7056 - 2(70)(84) \cdot 0.6157
   \]
   \[
   a^2 = 11956 - 7247.93
   \]
   \[
   a^2 = 4708.07 \quad \Rightarrow \quad a = \sqrt{4708.07} \approx 68.6 \, \text{mm}
   \]
   
2. **Law of Sines** to find the other angles:
   Using the Law of Sines:
   \[
   \frac{\sin B}{b} = \frac{\sin A}{a}
   \]
   \[
   \frac{\sin B}{70} = \frac{\sin 52^\circ}{68.6}
   \]
   \[
   \sin B = \frac{70 \cdot \sin 52^\circ}{68.6} = \frac{70 \cdot 0.7880}{68.6} \approx 0.804
   \]
   \[
   B \approx \sin^{-1}(0.804) \approx 53.4^\circ
   \]

   Finally, the third angle \( C \) is:
   \[
   C = 180^\circ - A - B = 180^\circ - 52^\circ - 53.4^\circ = 74.6^\circ
   \]

Thus, for Triangle A:
- \( a \approx 68.6 \, \text{mm} \)
- \( B \approx 53.4^\circ \)
- \( C \approx 74.6^\circ \)

---

### Triangle B:
- Given:
  - Angle \( A = 66^\circ \)
  - Angle \( C = 48^\circ \)
  - Side \( c = 14 \, \text{cm} \)

We can find the third angle and then use the **Law of Sines** to find the other sides.

1. **Third Angle**:
   \[
   B = 180^\circ - A - C = 180^\circ - 66^\circ - 48^\circ = 66^\circ
   \]

2. **Law of Sines** to find sides \( a \) and \( b \):
   \[
   \frac{a}{\sin A} = \frac{c}{\sin C}
   \]
   \[
   a = \frac{14 \cdot \sin 66^\circ}{\sin 48^\circ}
   \]
   \[
   a = \frac{14 \cdot 0.9135}{0.7431} \approx 17.2 \, \text{cm}
   \]
   
   Similarly, for side \( b \):
   \[
   \frac{b}{\sin B} = \frac{c}{\sin C}
   \]
   Since \( B = 66^\circ \), we can use the same result:
   \[
   b = \frac{14 \cdot \sin 66^\circ}{\sin 48^\circ} \approx 17.2 \, \text{cm}
   \]

Thus, for Triangle B:
- \( a \approx 17.2 \, \text{cm} \)
- \( b \approx 17.2 \, \text{cm} \)
- \( B = 66^\circ \)

---

### Triangle C:
- Given:
  - The sides are in terms of \( x \): \( a = 3x \), \( b = 4x \), \( c = 25 \, \text{m} \)
  
We can apply the **Law of Cosines** to find \( x \) using the given side lengths and angles.

Let's assume the largest angle is opposite the longest side (which would give us \( c = 25 \, \text{m} \)). Once we have more information, we can solve for \( x \).

---

Do you have any questions, or would you like details on any specific part?

### Follow-up Questions:
1. How do you apply the Law of Cosines in a triangle with two sides and an angle?
2. What is the significance of using the Law of Sines versus the Law of Cosines?
3. How can you use the sum of angles in a triangle to check your work?
4. In triangle C, how would you find the value of \( x \) once you have more data?
5. Why is the third angle in each triangle always calculated using \( 180^\circ \)?

### Tip:
Always remember to check if your triangle is right-angled before choosing between the Law of Sines and the Law of Cosines!

Use the given information to find all sides and angles in each triangle.

Solve triangles using the Law of Sines and Law of Cosines. This problem explores the trigonometric relationships between sides and angles, with detailed step-by-step solutions for each triangle. Suitable for high school students in Grades 10-12.

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