1. **Problem statement:** A telephone pole is supported by two wires forming a 70° angle at the top. The ends of the wires on the ground are 20 m apart. One wire makes a 30° angle with the ground. We need to find: a) The lengths of the wires. b) The height of the pole. 2. **Setup and known values:** - Angle between wires at the top: $70^\circ$ - Distance between wire ends on the ground: $20$ m - Angle of one wire with the ground: $30^\circ$ 3. **Diagram and variables:** Let the pole height be $h$. Let the lengths of the wires be $L_1$ (wire with 30° angle) and $L_2$ (other wire). The wires meet at the top of the pole forming a 70° angle. 4. **Find the angle of the other wire with the ground:** Since the wires form a 70° angle at the top, and one wire is at 30° with the ground, the other wire's angle with the ground is: $$\theta_2 = 70^\circ + 30^\circ = 100^\circ$$ 5. **Coordinates of wire ends on the ground:** Place the pole at origin $(0,0)$. The wire with 30° angle ends at: $$x_1 = L_1 \cos 30^\circ, \quad y_1 = 0$$ The other wire ends at: $$x_2 = -L_2 \cos 80^\circ, \quad y_2 = 0$$ Note: The second wire angle with ground is $100^\circ$, so its horizontal component is $\cos(180^\circ - 100^\circ) = -\cos 80^\circ$. 6. **Distance between wire ends on ground:** Given as 20 m, so: $$x_1 - x_2 = 20$$ Substitute: $$L_1 \cos 30^\circ - (-L_2 \cos 80^\circ) = 20$$ $$L_1 \cos 30^\circ + L_2 \cos 80^\circ = 20$$ 7. **Height of the pole:** Both wires reach the same height $h$: $$h = L_1 \sin 30^\circ = L_2 \sin 80^\circ$$ From this: $$L_1 = \frac{h}{\sin 30^\circ}, \quad L_2 = \frac{h}{\sin 80^\circ}$$ 8. **Substitute $L_1$ and $L_2$ into the distance equation:** $$\frac{h}{\sin 30^\circ} \cos 30^\circ + \frac{h}{\sin 80^\circ} \cos 80^\circ = 20$$ 9. **Simplify:** $$h \left( \frac{\cos 30^\circ}{\sin 30^\circ} + \frac{\cos 80^\circ}{\sin 80^\circ} \right) = 20$$ 10. **Calculate trigonometric values:** $$\cos 30^\circ = 0.8660, \quad \sin 30^\circ = 0.5$$ $$\cos 80^\circ = 0.1736, \quad \sin 80^\circ = 0.9848$$ 11. **Evaluate the expression:** $$h \left( \frac{0.8660}{0.5} + \frac{0.1736}{0.9848} \right) = 20$$ $$h (1.732 + 0.176) = 20$$ $$h (1.908) = 20$$ 12. **Solve for $h$:** $$h = \frac{20}{1.908} = 10.48 \approx 10.5 \text{ m}$$ 13. **Find wire lengths:** $$L_1 = \frac{h}{\sin 30^\circ} = \frac{10.48}{0.5} = 20.96 \approx 21.0 \text{ m}$$ $$L_2 = \frac{h}{\sin 80^\circ} = \frac{10.48}{0.9848} = 10.64 \approx 10.6 \text{ m}$$ **Final answers:** - a) Lengths of wires: $L_1 = 21.0$ m, $L_2 = 10.6$ m - b) Height of the pole: $h = 10.5$ m