A commercial airline pilot is flying at an altitude of 6.5 miles. To make a gentle descent for landing, the pilot begins descending toward the airport when still fairly far away. a) If the pilot begins descending 186 miles from the airport (measured on the ground), what angle will the plane’s path make with the runway (without further adjustment)? b) If the plane’s path is to make an angle of 5° with the runway (without further adjustment), how far from the airport (measured on the ground) must the pilot begin descending?

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RenatoMattice RenatoMattice · Answer 1 · 2019-01-14T07:55:53+01:00

Answer: a) the angle of elevation which plane's make with the runway is [tex]2\textdegree[/tex]

b) it takes 74.58 miles away from the airport must the pilot begin descending.

Step-by-step explanation:

Height of the commercial airline pilot is flying (PQ)= 6.5 miles

Length from which the pilot begins descending from the airport (PR)= 186 miles

We need to find the angle of elevation:

Consider, ΔPQR,

[tex]\sin \theta=\frac{PQ}{PR}\\\\\sin \theta=\frac{6.5}{186}\\\\\sin \theta=0.035\textdegree\\\\\theta=\sin^{-1}(0.035)\\\\\theta=2\textdegree[/tex]

Hence, the angle of elevation which plane's make with the runway is [tex]2\textdegree[/tex]

Now, According to question,

If the plane’s path is to make an angle of 5° with the runway:

Consider, ΔPQS,

[tex]\sin \theta=\frac{PQ}{PS}\\\\\sin \theta=\frac{6.5}{PS}\\\\\sin 5=\frac{6.5}{PS}\\\\PS=74.58[/tex]

Hence, it takes 74.58 miles away from the airport must the pilot begin descending.