Let's go ahead and solve each triangle in the image by finding the missing sides or angles.

### Triangle 1 (Top Left):
- Given: One angle is \(45^\circ\), and one side adjacent to the angle is 10 cm.
  
**Solution:**
- Since it's a \(45^\circ\) angle, the other non-right angle must also be \(45^\circ\) (because the angles in a triangle sum up to \(180^\circ\)).
- This means the triangle is an **isosceles right triangle**. Thus, the two legs are equal.
  
Using the Pythagorean theorem:
\[
c = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2} \approx 14.14 \text{ cm}
\]

### Triangle 2 (Top Middle):
- Given: One angle is \(30^\circ\), and the hypotenuse is 8 cm.

**Solution:**
- For a \(30^\circ\) angle, the opposite side is half of the hypotenuse:
\[
\text{Opposite} = \frac{8}{2} = 4 \text{ cm}
\]
- The adjacent side (using Pythagorean theorem):
\[
\text{Adjacent} = \sqrt{8^2 - 4^2} = \sqrt{64 - 16} = \sqrt{48} \approx 6.93 \text{ cm}
\]

### Triangle 3 (Top Right):
- Given: One angle is \(60^\circ\) and the side opposite this angle is 6.9 cm.

**Solution:**
- Using sine:
\[
\sin(60^\circ) = \frac{\text{Opposite}}{\text{Hypotenuse}} \Rightarrow \text{Hypotenuse} = \frac{6.9}{\sin(60^\circ)} \approx 7.96 \text{ cm}
\]
- For the adjacent side:
\[
\cos(60^\circ) = \frac{\text{Adjacent}}{\text{Hypotenuse}} \Rightarrow \text{Adjacent} = \cos(60^\circ) \times 7.96 \approx 3.98 \text{ cm}
\]

### Triangle 4 (Middle Left):
- Given: One side (adjacent to \(60^\circ\)) is 20 cm.

**Solution:**
- Hypotenuse using cosine:
\[
\cos(60^\circ) = \frac{\text{Adjacent}}{\text{Hypotenuse}} \Rightarrow \text{Hypotenuse} = \frac{20}{\cos(60^\circ)} = 40 \text{ cm}
\]
- Opposite side using sine:
\[
\sin(60^\circ) = \frac{\text{Opposite}}{\text{Hypotenuse}} \Rightarrow \text{Opposite} = \sin(60^\circ) \times 40 \approx 34.64 \text{ cm}
\]

### Triangle 5 (Middle Right):
- Given: Hypotenuse is 15 cm, one side is unknown.

**Solution:**
- If one angle is \(30^\circ\), then:
\[
\text{Opposite} = \frac{15}{2} = 7.5 \text{ cm}
\]
- Adjacent side:
\[
\cos(30^\circ) = \frac{\text{Adjacent}}{15} \Rightarrow \text{Adjacent} = \cos(30^\circ) \times 15 \approx 12.99 \text{ cm}
\]

### Triangle 6 (Bottom Left):
- Given: Side opposite \(30^\circ\) is 42 mm.

**Solution:**
- Hypotenuse (double the opposite side):
\[
\text{Hypotenuse} = 2 \times 42 = 84 \text{ mm}
\]
- Adjacent side:
\[
\cos(30^\circ) = \frac{\text{Adjacent}}{\text{Hypotenuse}} \Rightarrow \text{Adjacent} = \cos(30^\circ) \times 84 \approx 72.75 \text{ mm}
\]

### Triangle 7 (Bottom Right):
- Given: Angle \(22^\circ\), adjacent side 15 cm.

**Solution:**
- Hypotenuse using cosine:
\[
\cos(22^\circ) = \frac{15}{\text{Hypotenuse}} \Rightarrow \text{Hypotenuse} = \frac{15}{\cos(22^\circ)} \approx 16.16 \text{ cm}
\]
- Opposite side:
\[
\sin(22^\circ) = \frac{\text{Opposite}}{16.16} \Rightarrow \text{Opposite} = \sin(22^\circ) \times 16.16 \approx 6.06 \text{ cm}
\]

Would you like any further explanations or need more details on the steps used?

### Follow-up Questions:
1. How can you determine which trigonometric function to use given a specific side and angle?
2. What is the relationship between the angles and sides in a \(45^\circ-45^\circ-90^\circ\) triangle?
3. How can the sine and cosine functions be derived from the unit circle?
4. What is the role of complementary angles in determining the sides of a triangle?
5. How would you solve a right triangle if only one side and the right angle are known?

**Tip:** When solving right triangles, always cross-check your calculations by verifying that the sum of the squares of the two shorter sides equals the square of the hypotenuse.

Explore how to solve right triangle problems step-by-step with trigonometry. Includes detailed solutions for triangles with given angles and sides using the Pythagorean theorem, sine, and cosine functions. Suitable for high school students studying trigonometry.

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