Question

An airplane is at an elevation of 35,000 ft when it starts its descent at a 20 degree angle of depression. What is the air distance between the airplane and the airport in miles?

Answer 1

19.38 miles
This is a trigonometry problem. The angle of depression is 20 degrees and is the angle used with the trig ratio. Compared to that angle the altitude is the opposite side and the air distance is the hypotenuse. Change feet to miles by dividing 35000 by 5280 which is about 6.63 miles. opposite and hypotenuse is used in a sine function with the formula.
$sin \theta = \frac{opp}{hyp}$
plug in known values
$sin(20) = \frac{6.63}{x}$
Switch sin20 and x using the products property
$x = \frac{6.63}{sin(20)}$
plug into a calculator to get the answer of 19.38

Answer 2

Answer:

19.4 miles

Step-by-step explanation:

Let x represent distance between the airplane and the airport in miles.

We have been given that an airplane is at an elevation of 35,000 ft when it starts its descent at a 20 degree angle of depression. We are asked to find the distance between the airplane and the airport in miles.

We can see that airplane, airport and angle of depression forms a right triangle with respect to ground. The side of 35,000 ft is opposite side and x is hypotenuse to 20 degree angle.

$\text{sin}=\frac{\text{Opposite}}{\text{Hypotenuse}}$

$\text{sin}(20^{\circ})=\frac{35,000}{x}$

$x=\frac{35,000}{\text{sin}(20^{\circ})}$

$x=\frac{35,000}{0.342020143326}$

$x=102333.154$

To convert our distance into miles, we will divide 102333.154 by 5280.

$x=\frac{102333.154}{5280}$

$x=19.381279$

$x\approx 19.4$

Therefore, the air distance between the airplane and the airport is 19.4 miles.