Two airplanes leave the airport. plane a departs at a 42° angle from the runway, and plane b departs at a 44° from the runway. which plane was farther away from the airport when it was 15 miles from the ground? round the solutions to the nearest hundredth.
plane a bc it was 20.18 miles away
plane a bc it was 16.66 miles away
plane b bcit was 20.85 miles away
plane b bc it was 15.53 miles away

  Plane A bc it was 16.66 miles away

Step-by-step explanation:

The plane flying at the shallower angle will take more distance to achieve the same height. You know without any calculation that plane A will be farther away at the same height.

The mnemonic SOH CAH TOA reminds you that the relationship between the ground distance and the height is ...

  Tan = Opposite/Adjacent

  tan(42°) = (15 miles)/(ground distance) . . . . substitute given values

Solving for ground distance, we get ...

  ground distance = (15 miles)/tan(42°) ≈ 16.66 miles

Comment on the problem

Unless the problem is carefully worded (this one isn't), there is always some ambiguity as to what distance is being measured when you talk about distance from an airport. Here, we assumed the distance of interest is the distance measured along the ground to the point directly below the airplane. The straight-line distance from the plane to the airport is about ...

  (15 miles)/sin(42°) ≈ 22.42 miles

Since this is not an answer choice, we assume that the ground distance is what we're concerned with.

first one

step-by-step explanation: