wo 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 7 miles from the ground? round the solutions to the nearest hundredth.
plane a because it was 9.428 miles away
plane a because it was 10.46 miles away
plane b because it was 10.08 miles away
plane b because it was 9.73 miles away

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Answer:

plane a because it was 10.46 miles away

Step-by-step explanation:

We use the Sine Rule to determine which plane is further away after 7 miles off the ground.

#Plane A departs at 42°, The 7 mile flight corresponds to the 42°:

[tex]\frac{a}{Sin \ A}=\frac{b}{Sin \ B}\\\\\frac{7}{Sin \ 42\textdegree}=\frac{a}{Sin \ 90\textdegree}\\\\a=7\ Sin \ 42\textdegree\\\\=10.46\ miles[/tex]

#Plane B departs at  44°,The 7 mile flight corresponds to the 42°:

[tex]\frac{a}{Sin \ A}=\frac{b}{Sin \ B}\\\\\frac{7}{Sin \ 44\textdegree}=\frac{b}{Sin \ 90\textdegree}\\\\b=7\ Sin \ 44\textdegree\\\\=10.08\ miles[/tex]

Hence, Plane A since it's 10.46 miles away.

Answer:

plane A 10.46

Step-by-step explanation:

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