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A plane is at a cruising altitude of 10,150 feet. If the angle of depression to the runway is 23 degrees, what is the distance, to the nearest foot, from the planes cruising altitude to landing on the runway?

Answer :

Answer:

The distance to the nearest foot, from the plane altitude is 25979 feet

Step-by-step explanation:

Given as :

The altitude at which the plane is = 10,150 feet

The angle of depression on the runway = 23°

Let The distance to the nearest foot, from the plane altitude = x feet

Now, From figure

Sin angle = [tex]\dfrac{\textrm perpendicular}{\textrm hypotenuse}[/tex]

Or, Sin 23° =  [tex]\dfrac{\textrm AB }{\textrm BO}[/tex]

Or, 0.3907 =  [tex]\dfrac{\textrm 10,150 }{\textrm x}[/tex]

∴ x =  [tex]\dfrac{\textrm 10,150 }{\textrm 0.3907}[/tex]

Or, x = 25979

I.e  The distance to the nearest foot, from the plane altitude = x = 25979 feet

Hence The distance to the nearest foot, from the plane altitude is 25979 feet . Answer

${teks-lihat-gambar} WaywardDelaney

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