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.