Question

an airplane 8 miles from the airport is descending at ar angle of 6° with the ground. when the plane reaches 2. miles from the airport, the angle of descent becomes 3 with the ground as shown in the diagram below. which is closest to the height of the airplane above th round at point a, when the angle of descent changes 3.5°? (note: 1 mile = 5280 feet) 807 ft 1776 ft 1271 ft 2796 ft

Explanation:

Step1: Use the trigonometric ratio for the second - stage descent

The angle of descent is \(3.5^{\circ}\), and the horizontal distance from the airport is \(2.5\) miles. We know that \(\tan\theta=\frac{h}{d}\), where \(h\) is the height and \(d\) is the horizontal distance.
So, \(h = d\times\tan\theta\).

Step2: Substitute the values

Substitute \(d = 2.5\) miles and \(\theta=3.5^{\circ}\) into the formula. First, convert miles to feet. Since \(1\) mile \( = 5280\) feet, \(d=2.5\times5280 = 13200\) feet.
And \(\tan(3.5^{\circ})\approx0.06116\) (using a calculator, \(\tan(x)\) where \(x\) is in degrees, \(\tan(3.5^{\circ})=\frac{\sin(3.5^{\circ})}{\cos(3.5^{\circ})}\approx0.06116\)).
Then \(h=13200\times0.06116\).

Step3: Calculate the value of \(h\)

\(h = 13200\times0.06116=807.312\) feet.

Answer:

807 ft