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!

answer: yes, he is correct

step-by-step explanation: 10x-1=59

                                              1-1=0     59+1=60

60/10=x=6

f

step-by-step explanation:

f