1. **Problem statement:** Two cars start from the same intersection where roads meet at 34°. The slower car travels at 80 km/h, the faster at 100 km/h. After 2 hours, a helicopter above and between them measures a 20° angle of depression to the slower car and a straight-line distance of 100 km to it. We need to find: a) The straight-line distance from the helicopter to the faster car. b) The altitude of the helicopter. 2. **Known values:** - Angle between roads: $34^\circ$ - Speed slower car: $80$ km/h - Speed faster car: $100$ km/h - Time traveled: $2$ h - Distance helicopter to slower car: $100$ km - Angle of depression to slower car: $20^\circ$ 3. **Calculate distances traveled by cars:** - Slower car distance: $80 \times 2 = 160$ km - Faster car distance: $100 \times 2 = 200$ km 4. **Set up coordinate system:** Place the intersection at origin $O$. Let the slower car be at point $A$ along one road, and the faster car at point $B$ along the other road, with angle $\angle AOB = 34^\circ$. 5. **Helicopter position:** Helicopter $H$ is above the plane between $A$ and $B$. The angle of depression to $A$ is $20^\circ$, and $HA = 100$ km (straight-line distance). 6. **Find horizontal distance from helicopter to slower car:** Using angle of depression, $$\sin 20^\circ = \frac{\text{altitude } h}{100} \implies h = 100 \sin 20^\circ$$ Calculate altitude: $$h = 100 \times 0.3420 = 34.20 \text{ km}$$ 7. **Find horizontal distance from helicopter to slower car's ground point:** $$d = 100 \cos 20^\circ = 100 \times 0.9397 = 93.97 \text{ km}$$ 8. **Coordinates of points:** - $A$ at $(160,0)$ (along x-axis) - $B$ at $(200 \cos 34^\circ, 200 \sin 34^\circ)$ Calculate $B$: $$200 \times \cos 34^\circ = 200 \times 0.8290 = 165.8$$ $$200 \times \sin 34^\circ = 200 \times 0.5592 = 111.8$$ So $B = (165.8, 111.8)$ 9. **Helicopter $H$ lies on the line between $A$ and $B$ at horizontal distance $d=93.97$ km from $A$:** Vector $\overrightarrow{AB} = (165.8 - 160, 111.8 - 0) = (5.8, 111.8)$ Length $|AB| = \sqrt{5.8^2 + 111.8^2} \approx 111.95$ km Unit vector $\hat{u} = \left(\frac{5.8}{111.95}, \frac{111.8}{111.95}\right) = (0.0518, 0.9987)$ Position of $H$: $$H = A + t \hat{u}$$ where $t$ is the distance along $AB$ from $A$ to $H$. Since $H$ is $d=93.97$ km from $A$ horizontally, $$t = 93.97$$ Coordinates of $H$: $$x_H = 160 + 93.97 \times 0.0518 = 160 + 4.87 = 164.87$$ $$y_H = 0 + 93.97 \times 0.9987 = 93.85$$ 10. **Calculate straight-line distance from helicopter to faster car $B$:** Horizontal distance: $$\sqrt{(165.8 - 164.87)^2 + (111.8 - 93.85)^2} = \sqrt{0.93^2 + 17.95^2} = \sqrt{0.8649 + 322.2} = \sqrt{323.06} = 17.97 \text{ km}$$ Vertical difference is altitude $h=34.20$ km, so total distance: $$HB = \sqrt{17.97^2 + 34.20^2} = \sqrt{323.06 + 1169.64} = \sqrt{1492.7} = 38.64 \text{ km}$$ Rounded to nearest km: $$39 \text{ km}$$ 11. **Altitude of helicopter:** Already found as $34.20$ km, rounded to nearest km: $$34 \text{ km}$$ **Final answers:** - a) Distance from helicopter to faster car: **39 km** - b) Altitude of helicopter: **34 km**