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. W

Question

4 years ago

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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. W

hich plane was farther away from the airport when it was 18 miles from the ground? Round the solutions to the nearest hundredth. Plane A because it was 20.71 miles away Plane A because it was 27.43 miles away Plane B because it was 16.79 miles away Plane B because it was 26.39 miles away

Mathematics

2 answers:

alina1380 [7] · Answer 1 · 2022-01-10T12:46:41+03:00

alina1380 [7]4 years ago

7 0

Answer:

Plane A because it was 7.62 miles away

Step-by-step explanation:

took the test

lions [1.4K] · Answer 2 · 2022-01-10T10:44:07+03:00

Answer:

The correct option is Plane A because it was 27.43 miles away ....

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

then 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

The correct option is Plane A because it was 27.43 miles away ....