An airplane is flying at an altitude of 10,000 feet. the pilot notices that the beginning of the runway where he plans to land is at an angle of depression of 13 degrees with the horizontal. how far horizontally must he fly to be directly over the runway?

43314.76 ft horizontally he must fly to be directly over the runway

Step-by-step explanation:

Refer the figure given

We have altitude of aeroplane = BC = 10000 ft

The pilot notices that the beginning of the runway where he plans to land is at an angle of depression of 13 degrees with the horizontal

             ∠CAB = 13°

We need to find how far horizontally must he fly to be directly over the runway = AB

We have

43314.76 ft horizontally he must fly to be directly over the runway

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.