Find missing side lengths in triangles using trigonometry.

To solve the problem of finding the missing sides in the given triangles, we need to use trigonometric relationships such as sine, cosine, and tangent. Let's go through each part step by step.

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Problem 1:

#### Triangle with sides 10 and an angle of 30°
We are asked to find the length of side \( DF \).

- The triangle is a right triangle.
- One angle is 30°, and the hypotenuse is 10.
- We can use the sine function:
\[
\sin(30^\circ) = \frac{\text{opposite}}{\text{hypotenuse}}
\]
Here, the opposite side to the 30° angle is \( DF \), and the hypotenuse is 10.
\[
\sin(30^\circ) = \frac{DF}{10}
\]
We know that \(\sin(30^\circ) = \frac{1}{2}\).
\[
\frac{1}{2} = \frac{DF}{10}
\]
Solving for \( DF \):
\[
DF = 10 \times \frac{1}{2} = 5
\]

So, the length of \( DF \) is:
\[
\boxed{5}
\]

---

Problem 2:

#### Triangle with sides 8 and an angle of 45°
We are asked to find the length of side \( AC \).

- The triangle is a right triangle.
- One angle is 45°, and one leg is 8.
- Since it's a 45-45-90 triangle, both legs are equal.
\[
AC = 8
\]

So, the length of \( AC \) is:
\[
\boxed{8}
\]

---

Problem 3:

#### Triangle with sides 6 and an angle of 60°
We are asked to find the length of side \( KZ \).

- The triangle is a right triangle.
- One angle is 60°, and the adjacent side is 6.
- We can use the tangent function:
\[
\tan(60^\circ) = \frac{\text{opposite}}{\text{adjacent}}
\]
Here, the opposite side to the 60° angle is \( KZ \), and the adjacent side is 6.
\[
\tan(60^\circ) = \frac{KZ}{6}
\]
We know that \(\tan(60^\circ) = \sqrt{3}\).
\[
\sqrt{3} = \frac{KZ}{6}
\]
Solving for \( KZ \):
\[
KZ = 6 \times \sqrt{3} = 6\sqrt{3}
\]

So, the length of \( KZ \) is:
\[
\boxed{6\sqrt{3}}
\]

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Problem 4:

#### Triangle with sides 9 and an angle of 30°
We are asked to find the length of side \( TV \).

- The triangle is a right triangle.
- One angle is 30°, and the opposite side is 9.
- We can use the sine function:
\[
\sin(30^\circ) = \frac{\text{opposite}}{\text{hypotenuse}}
\]
Here, the opposite side to the 30° angle is 9, and the hypotenuse is \( TV \).
\[
\sin(30^\circ) = \frac{9}{TV}
\]
We know that \(\sin(30^\circ) = \frac{1}{2}\).
\[
\frac{1}{2} = \frac{9}{TV}
\]
Solving for \( TV \):
\[
TV = 9 \times 2 = 18
\]

So, the length of \( TV \) is:
\[
\boxed{18}
\]

---

Problem 5:

#### Triangle with sides 7 and an angle of 45°
We are asked to find the length of side \( CD \).

- The triangle is a right triangle.
- One angle is 45°, and the hypotenuse is 7.
- In a 45-45-90 triangle, the legs are equal, and each leg is \(\frac{\text{hypotenuse}}{\sqrt{2}}\).
\[
CD = \frac{7}{\sqrt{2}} = \frac{7\sqrt{2}}{2}
\]

So, the length of \( CD \) is:
\[
\boxed{\frac{7\sqrt{2}}{2}}
\]

---

Problem 6:

#### Triangle with sides 12 and an angle of 60°
We are asked to find the length of side \( XY \).

- The triangle is a right triangle.
- One angle is 60°, and the adjacent side is 12.
- We can use the tangent function:
\[
\tan(60^\circ) = \frac{\text{opposite}}{\text{adjacent}}
\]
Here, the opposite side to the 60° angle is \( XY \), and the adjacent side is 12.
\[
\tan(60^\circ) = \frac{XY}{12}
\]
We know that \(\tan(60^\circ) = \sqrt{3}\).
\[
\sqrt{3} = \frac{XY}{12}
\]
Solving for \( XY \):
\[
XY = 12 \times \sqrt{3} = 12\sqrt{3}
\]

So, the length of \( XY \) is:
\[
\boxed{12\sqrt{3}}
\]

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Final Answers:

1. \( DF = \boxed{5} \)
2. \( AC = \boxed{8} \)
3. \( KZ = \boxed{6\sqrt{3}} \)
4. \( TV = \boxed{18} \)
5. \( CD = \boxed{\frac{7\sqrt{2}}{2}} \)
6. \( XY = \boxed{12\sqrt{3}} \)

\boxed{5, 8, 6\sqrt{3}, 18, \frac{7\sqrt{2}}{2}, 12\sqrt{3}}