A plane takes off from an airport and flies at a speed of 400km/h on a course of 120° for 2 hours. the plane then changes its course to 210° and continues in this direction for 1 hour. How far is the plane from the airport at the end of this time?
Respuestas
Espanish
La distancia que se encutra el avion del aeropuerto es de
D = 894.43 Km
Explicación paso a paso:
Datos del problema:
- v = 400 km/h
- 120° por 2 horas
- 210° por 1 hora
Si se eleva a un angulo de 120° por 2 horas recorre una distancia de
400km/h * 2h = 800km
luego cambia su rumbo a 210° (medida desde la misma de despeje) por 1 hora
400km recorre
de 120° a 210° hay 90° el cambio de rumbo forma un angulo recto
En total recorre 1200km y la distancia directa es de
Teorema de pitagoras:
D = √(400km)²+(800km)²
D = 894.43 Km
English
The distance that the plane is from the airport is
D = 894.43 Km
Step by step explanation:
Problem data:
- v = 400 km / h
- 120 ° for 2 hours
- 210 ° for 1 hour
If it rises at an angle of 120 ° for 2 hours, it travels a distance of
400km / h * 2h = 800km
then change course to 210 ° (measured from clearing) for 1 hour
400km travels
from 120 ° to 210 ° there is 90 ° the change of course forms a right angle
In total it travels 1200km and the direct distance is
Pythagoras theorem:
D = √ (400km) ² + (800km) ²
D = 894.43 Km