Question

name: anthony adams mrs. morganunova, teacher yonkers montessori academy date: lesson 4: trigonometry review. word problems. challenge. an airplane traveling at a level altitude of 35000 feet above the ocean level. at point a the pilot saw a cruise ship #1. the angle of depression to the top of the ship was 72°. the height of the cruise ship was 70 meters. the airplane continues to travel. at point b (3000 feet from point a) the pilot saw a cruise ship #2. the angle of depression to the top of the ship was 54° and the height of the cruise ship was 76 meters. when the airplane was exactly above the cruise ship #1 it changed altitude by going down 2500 meters and saw the yacht was 328 ft. the angle of depression to the top of the yacht was 32° and height of the yacht was (from point c). the pilot saw a lighthouse. the angle of depression to the top of the lighthouse was 13° and the height of the lighthouse was 63 meters. a) draw a diagram that shows all information above (hint: create the picture step by step and do not draw next step until you finished with the previous one and labeled all information). b) find the horizontal distance first and second cruise ship in feet and then in meters (to the nearest whole number). c) find the horizontal distance from the yacht to the cruise ship #1 in meters (to the nearest whole number). d) find the horizontal distance from the lighthouse to the cruise ship #2 in meters (to the nearest whole number). e) find the horizontal distance from the lighthouse to the point a in meters (to the nearest whole number). 1 meter = 3.28084 feet

Explanation:

Step1: Identify relevant trigonometric relationships

We will use tangent function in right - angled triangles formed by height differences and horizontal distances. The tangent of the angle of depression is equal to the ratio of the height difference to the horizontal distance.

Step2: For part b (distance between first and second cruise ship)

Let's first consider the height difference between the two points of observation. The height of the airplane at point A is 35000 feet and at point B is \(35000 - 3000=32000\) feet. The height of cruise - ship #1 is \(h_1\) and of cruise - ship #2 is \(h_2\).
Let the horizontal distance between the two ships be \(d\).
For the angle of depression from point A to cruise - ship #1 (\(\theta_1 = 72^{\circ}\)) and from point B to cruise - ship #2 (\(\theta_2 = 54^{\circ}\)).
We know that \(\tan\theta=\frac{\text{height difference}}{\text{horizontal distance}}\).
Let the height of cruise - ship #1 be \(h_1\) (in feet) and of cruise - ship #2 be \(h_2\) (in feet). First, convert the heights of the ships from meters to feet: \(h_1 = 70\times3.28084\approx229.66\) feet and \(h_2 = 76\times3.28084\approx249.35\) feet.
The height of the airplane above cruise - ship #1 at point A is \(H_1 = 35000 - 229.66=34770.34\) feet.
The height of the airplane above cruise - ship #2 at point B is \(H_2 = 32000 - 249.35 = 31750.65\) feet.
Using \(\tan72^{\circ}=\frac{34770.34}{x_1}\) and \(\tan54^{\circ}=\frac{31750.65}{x_2}\), where \(x_1\) and \(x_2\) are the horizontal distances from the airplane to the respective ships at points A and B.
\(x_1=\frac{34770.34}{\tan72^{\circ}}\) and \(x_2=\frac{31750.65}{\tan54^{\circ}}\).
The horizontal distance between the two ships \(d = x_1 - x_2\).
\(\tan72^{\circ}\approx3.0777\), \(x_1=\frac{34770.34}{3.0777}\approx11300\) feet.
\(\tan54^{\circ}\approx1.3764\), \(x_2=\frac{31750.65}{1.3764}\approx23067\) feet.
\(d=x_2 - x_1=23067 - 11300 = 11767\) feet.
Convert to meters: \(d = 11767\div3.28084\approx3587\) meters.

Step3: For part c (distance from yacht to cruise - ship #1)

The height of the airplane above the yacht at point C is \(H_y=35000 - 2500-328\times3.28084\div1 = 35000 - 2500 - 1076.11=31423.89\) feet.
The height of the airplane above cruise - ship #1 is \(35000 - 70\times3.28084=35000 - 229.66 = 34770.34\) feet.
The angle of depression from the airplane above the yacht to the top of cruise - ship #1 is \(32^{\circ}\).
Using \(\tan32^{\circ}=\frac{\text{height difference}}{\text{horizontal distance}}\).
The height difference \(h = 34770.34-31423.89 = 3346.45\) feet.
\(\tan32^{\circ}\approx0.6249\), so the horizontal distance \(x=\frac{3346.45}{0.6249}\approx5355\) feet.
Convert to meters: \(x = 5355\div3.28084\approx1632\) meters.

Step4: For part d (distance from lighthouse to cruise - ship #2)

The height of the airplane above the lighthouse at point C is \(35000 - 2500-63\times3.28084=35000 - 2500 - 206.69 = 32293.31\) feet.
The height of the airplane above cruise - ship #2 is \(32000 - 76\times3.28084=32000 - 249.35 = 31750.65\) feet.
The angle of depression from the airplane above the lighthouse to the top of cruise - ship #2 is \(13^{\circ}\).
The height difference \(h = 32293.31 - 31750.65=542.66\) feet.
Using \(\tan13^{\circ}\approx0.2309\), the horizontal distance \(x=\frac{542.66}{0.2309}\approx2359\) feet.
Convert to meters: \(x = 2359\div3.28084\approx720\) meters.

Step5: For part e (distance from lighthouse to point A)

The height of the airplane at point A is 35000 feet and the height of the lighthouse is \(63\times3.28084 = 206.69\) feet.
The height difference \(h = 35000 - 206.69=34793.31\) feet.
The angle of depression from point A to the top of the lighthouse is \(13^{\circ}\).
Using \(\tan13^{\circ}\approx0.2309\), the horizontal distance \(x=\frac{34793.31}{0.2309}\approx150600\) feet.
Convert to meters: \(x = 150600\div3.28084\approx45900\) meters.

Answer:

b) In feet: 11767 feet; In meters: 3587 meters
c) 1632 meters
d) 720 meters
e) 45900 meters