Question

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. at point b (3000 feet above point a) the pilot saw a cruise ship #2. the angle of depression to the top of the ship was 54°. the height of the cruise ship was 76 meters. when the airplane was exactly above the cruise ship #1 it changed altitude going down 2500 meters and saw the yacht (from point c). the angle of depression to the top of the yacht was 32° and height of the yacht was 328 ft. from the same 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 from the 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: Convert feet to meters

We know that 1 meter = 3.28084 feet. At point A, the height of the airplane is 35000 feet, so in meters it is $h_A=\frac{35000}{3.28084}\approx10668$ meters. At point B, the height of the airplane is 30000 feet, so in meters it is $h_B = \frac{30000}{3.28084}\approx9144$ meters. When it goes down 2500 meters from point B, the height at point C is $h_C=9144 - 2500=6644$ meters. The height of cruise - ship #1 is 70 meters, of cruise - ship #2 is 76 meters and of the yacht is 328 feet which is $\frac{328}{3.28084}\approx100$ meters and the height of the lighthouse is 63 meters.

Step2: Use tangent function for horizontal distance (for part b)

Let the horizontal distance between the first and the second cruise - ship be $d_{12}$. We know that for the angle of depression from the airplane at point A to cruise - ship #1 is 72° and to cruise - ship #2 is 54°. Let the horizontal distance from the airplane at point A to cruise - ship #1 be $x_1$ and to cruise - ship #2 be $x_2$. We use the tangent function $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. For cruise - ship #1, $\tan72^{\circ}=\frac{h_A - 70}{x_1}$, so $x_1=\frac{10668 - 70}{\tan72^{\circ}}\approx3300$ meters. For cruise - ship #2, $\tan54^{\circ}=\frac{h_A - 76}{x_2}$, so $x_2=\frac{10668 - 76}{\tan54^{\circ}}\approx7700$ meters. Then $d_{12}=x_2 - x_1\approx7700-3300 = 4400$ meters.

Step3: Use tangent function for horizontal distance (for part c)

Let the horizontal distance from the yacht to cruise - ship #1 be $d_{yc1}$. The height difference between the airplane at point C and the yacht is $h_{C - y}=6644 - 100 = 6544$ meters and the angle of depression to the yacht is 32°. Using $\tan32^{\circ}=\frac{h_{C - y}}{d_{yc1}}$, we get $d_{yc1}=\frac{6544}{\tan32^{\circ}}\approx10470$ meters. The height difference between the airplane at point C and cruise - ship #1 is $h_{C - 1}=6644 - 70=6574$ meters. Using $\tan32^{\circ}=\frac{h_{C - 1}}{d_{C1}}$, we get $d_{C1}=\frac{6574}{\tan32^{\circ}}\approx10590$ meters. The horizontal distance from the yacht to cruise - ship #1 is $d_{yc1}-d_{C1}\approx10470 - 10590=- 120$ (this is wrong, we should use the height difference directly). The height difference between the yacht and cruise - ship #1 is $100 - 70 = 30$ meters. Let the horizontal distance be $d$. $\tan32^{\circ}=\frac{30}{d}$, so $d=\frac{30}{\tan32^{\circ}}\approx48$ meters.

Step4: Use tangent function for horizontal distance (for part d)

Let the horizontal distance from the lighthouse to cruise - ship #2 be $d_{l2}$. The height difference between the airplane at point C and the lighthouse is $h_{C - l}=6644 - 63=6581$ meters. The angle of depression to the lighthouse is 13°. Using $\tan13^{\circ}=\frac{h_{C - l}}{d_{l2}}$, we get $d_{l2}=\frac{6581}{\tan13^{\circ}}\approx28700$ meters. The height difference between the airplane at point C and cruise - ship #2 is $h_{C - 2}=6644 - 76 = 6568$ meters. Using $\tan13^{\circ}=\frac{h_{C - 2}}{d_{C2}}$, we get $d_{C2}=\frac{6568}{\tan13^{\circ}}\approx28600$ meters. The horizontal distance from the lighthouse to cruise - ship #2 is $d_{l2}-d_{C2}\approx100$ meters.

Step5: Use tangent function for horizontal distance (for part e)

Let the horizontal distance from the lighthouse to point A be $d_{lA}$. The height difference between the airplane at point A and the lighthouse is $h_{A - l}=10668 - 63 = 10605$ meters. The angle of depression from point A to the lighthouse is 13°. Using $\tan13^{\circ}=\frac{h_{A - l}}{d_{lA}}$, we get $d_{lA}=\frac{10605}{\tan13^{\circ}}\approx46000$ mete…

Answer:

b. 4400 meters
c. 48 meters
d. 100 meters
e. 46000 meters