Set of nine numbered triangles used for geometry exercises, featuring side lengths of 12cm and x with a 40-degree angle.

Problem Analysis

The image contains nine triangles, each with a labeled side and an angle of \(40^\circ\). The goal is to solve for the unknown side \(x\) in each triangle. We will use trigonometric relationships (sine, cosine, or tangent) to find \(x\) in each case.

Key Trigonometric Relationships

1. Sine Rule: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\)
2. Cosine Rule: \(c^2 = a^2 + b^2 - 2ab \cos C\)
3. Tangent Rule: \(\tan \theta = \frac{\text{opposite}}{\text{adjacent}}\)

Solution for Each Triangle

#### Triangle 1
- Given: \(12 \, \text{cm}\) (hypotenuse), \(40^\circ\), and \(x\) (opposite side).
- Use the sine function: \(\sin 40^\circ = \frac{x}{12}\)
- Solve for \(x\):
\[
x = 12 \cdot \sin 40^\circ
\]
Using \(\sin 40^\circ \approx 0.6428\):
\[
x \approx 12 \cdot 0.6428 = 7.7136 \, \text{cm}
\]

#### Triangle 2
- Given: \(12 \, \text{cm}\) (one leg), \(40^\circ\), and \(x\) (hypotenuse).
- Use the cosine function: \(\cos 40^\circ = \frac{12}{x}\)
- Solve for \(x\):
\[
x = \frac{12}{\cos 40^\circ}
\]
Using \(\cos 40^\circ \approx 0.7660\):
\[
x \approx \frac{12}{0.7660} \approx 15.665 \, \text{cm}
\]

#### Triangle 3
- Given: \(12 \, \text{cm}\) (one leg), \(40^\circ\), and \(x\) (adjacent side).
- Use the tangent function: \(\tan 40^\circ = \frac{12}{x}\)
- Solve for \(x\):
\[
x = \frac{12}{\tan 40^\circ}
\]
Using \(\tan 40^\circ \approx 0.8391\):
\[
x \approx \frac{12}{0.8391} \approx 14.299 \, \text{cm}
\]

#### Triangle 4
- Given: \(12 \, \text{cm}\) (one side), \(40^\circ\), and \(x\) (another side in an isosceles triangle).
- Use the Law of Sines: \(\frac{x}{\sin 40^\circ} = \frac{12}{\sin 100^\circ}\)
- Since the triangle is isosceles, the other angle is also \(40^\circ\), so the third angle is \(100^\circ\).
- Solve for \(x\):
\[
x = 12 \cdot \frac{\sin 40^\circ}{\sin 100^\circ}
\]
Using \(\sin 40^\circ \approx 0.6428\) and \(\sin 100^\circ \approx 0.9848\):
\[
x \approx 12 \cdot \frac{0.6428}{0.9848} \approx 12 \cdot 0.6528 \approx 7.8336 \, \text{cm}
\]

#### Triangle 5
- Given: \(12 \, \text{cm}\) (one side), \(40^\circ\), and \(x\) (hypotenuse in a right triangle).
- Use the cosine function: \(\cos 40^\circ = \frac{12}{x}\)
- Solve for \(x\):
\[
x = \frac{12}{\cos 40^\circ}
\]
Using \(\cos 40^\circ \approx 0.7660\):
\[
x \approx \frac{12}{0.7660} \approx 15.665 \, \text{cm}
\]

#### Triangle 6
- Given: \(12 \, \text{cm}\) (one side), \(40^\circ\), and \(x\) (adjacent side in a right triangle).
- Use the tangent function: \(\tan 40^\circ = \frac{12}{x}\)
- Solve for \(x\):
\[
x = \frac{12}{\tan 40^\circ}
\]
Using \(\tan 40^\circ \approx 0.8391\):
\[
x \approx \frac{12}{0.8391} \approx 14.299 \, \text{cm}
\]

#### Triangle 7
- Given: \(12 \, \text{cm}\) (one side), \(40^\circ\), and \(x\) (opposite side in a right triangle).
- Use the sine function: \(\sin 40^\circ = \frac{x}{12}\)
- Solve for \(x\):
\[
x = 12 \cdot \sin 40^\circ
\]
Using \(\sin 40^\circ \approx 0.6428\):
\[
x \approx 12 \cdot 0.6428 = 7.7136 \, \text{cm}
\]

#### Triangle 8
- Given: \(12 \, \text{cm}\) (one side), \(40^\circ\), and \(x\) (adjacent side in a right triangle).
- Use the tangent function: \(\tan 40^\circ = \frac{12}{x}\)
- Solve for \(x\):
\[
x = \frac{12}{\tan 40^\circ}
\]
Using \(\tan 40^\circ \approx 0.8391\):
\[
x \approx \frac{12}{0.8391} \approx 14.299 \, \text{cm}
\]

#### Triangle 9
- Given: \(14 \, \text{cm}\) (one side), \(12 \, \text{cm}\) (another side), and \(x\) (third side).
- Use the Law of Cosines: \(x^2 = 14^2 + 12^2 - 2 \cdot 14 \cdot 12 \cdot \cos 40^\circ\)
- Solve for \(x\):
\[
x^2 = 196 + 144 - 2 \cdot 14 \cdot 12 \cdot \cos 40^\circ
\]
Using \(\cos 40^\circ \approx 0.7660\):
\[
x^2 = 196 + 144 - 2 \cdot 14 \cdot 12 \cdot 0.7660
\]
\[
x^2 = 340 - 257.376
\]
\[
x^2 \approx 82.624
\]
\[
x \approx \sqrt{82.624} \approx 9.09 \, \text{cm}
\]

Final Answers

\[
\boxed{
\begin{aligned}
1. & \, 7.71 \, \text{cm} \\
2. & \, 15.67 \, \text{cm} \\
3. & \, 14.30 \, \text{cm} \\
4. & \, 7.83 \, \text{cm} \\
5. & \, 15.67 \, \text{cm} \\
6. & \, 14.30 \, \text{cm} \\
7. & \, 7.71 \, \text{cm} \\
8. & \, 14.30 \, \text{cm} \\
9. & \, 9.09 \, \text{cm}
\end{aligned}
}
\]