Respuesta :
Based on the given problem above, it requires a trigonometric solution. So here it goes:
The climb distance is:200 * Sec [ArcTan [10/200]] = 10√ 401 km
and the decent distance is:300 * Sec [ArcTan [10/300]] = 10√901 km
So add the given results above with 500 km and this will be the additional distance that plane moves through the air.
The answer would be in the unit meters.
((10√401 +10√901−500)∗1000)=416 meters
Hope this is the answer that you are looking for.
The climb distance is:200 * Sec [ArcTan [10/200]] = 10√ 401 km
and the decent distance is:300 * Sec [ArcTan [10/300]] = 10√901 km
So add the given results above with 500 km and this will be the additional distance that plane moves through the air.
The answer would be in the unit meters.
((10√401 +10√901−500)∗1000)=416 meters
Hope this is the answer that you are looking for.
Answer:
Climbing and descending add to the flight 416 m
Step-by-step explanation:
The plane must climb to 10 km during the first 200 km. This forms a right triangle, where 10 and 200 are its legs and the hypotenuse is the distance travelled by the airplane. From Pythagorean theorem, that distance is:
√(10² + 200²)
The plane must descend 10 km during the final 300km. This forms a right triangle, where 10 and 300 are its legs and the hypotenuse is the distance travelled by the airplane. From Pythagorean theorem, that distance is:
√(10² + 300²)
If the airplane climbed during 300 km and descended during 200 km, then during 1500 - 200 - 300 = 1000 km it doesn't climb or descend.
The distance added by climbing and descending is:
1000 + √(10² + 200²) + √(10² + 300²) - 1500 = 0.416 km or 416 m