Two airplanes leave the airport. plane a departs at a 41° angle from the runway, and plane b departs at a 43° from the runway. which plane was farther away from the airport when it was 18 miles from the ground? round the solutions to the nearest hundredth.

Plane A

Step-by-step explanation:

angle of elevation of plane A=41 °

angle of elevation of plane B=43°

height of planes=18 miles

let Δabc be the right triangle formed by the plane A and Δabd be the right triangle formed by the plane B

the ab=18miles

∠c, angle of elevation of plane A=41 °

then distance bc=?

sinc=ab/bc

sin(41°)=18/bc

bc=18/0.65

bc=27.437 miles

Now

∠d, angle of elevation of plane B=43°

then distance bd=?

sinc=ab/bd

sin(43°)=18/bd

bd=18/0.68199

bd=26.393 miles

As bc>bd

Hence, Plane A was farther away from the airport when it was 18 miles from the ground!

it could 14.28 but if its not that one then it has to be the 17 one. hope this ! : )

step-by-step explanation:

step-by-step :