An airplane climbs at an angle of 2.7" from an altitude of 4500 ft to an altitude of 5600 ft. How far does the airplane travel as it climbs as measured along the horizontal?

  23,325 ft, about 4.42 miles

Step-by-step explanation:

The geometry of the situation can be modeled by a right triangle, where the side adjacent to the 2.7° angle of climb is the horizontal distance, and the side opposite is the vertical change in altitude. That change is ...

  5600 ft -4500 ft = 1100 ft

and the angle relation is ...

  tan(2.7°) = opposite/adjacent = (1100 ft)/(horizontal distance)

Multiplying this equation by (horizontal distance)/tan(2.7°) gives ...

   horizontal distance = (1100 ft)/tan(2.7°) ≈ 23,325 ft

Dividing this by 5280 ft/mi gives ...

  horizontal distance ≈ 4.42 mi

The airplane travels about 23,325 ft, or 4.42 miles, horizontally as it climbs.

< 4, -5>

step-by-step explanation:

starting from g. we move 4 right and then 5 down to get to h. that means the vector is < 4, -5>