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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 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 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
Step1: Convert units and set up for ship 1
The altitude of the airplane above the ocean level is $35000$ feet. The height of cruise - ship 1 is $70$ meters. First, convert the height of the ship to feet: $70\times3.28084 = 229.6588\approx230$ feet. The vertical distance from the airplane to the top of the ship is $h_1=35000 - 230=34770$ feet. The angle of depression to the top of ship 1 is $\theta_1 = 72^{\circ}$. Since the angle of depression is equal to the angle of elevation from the ship to the airplane, and we know that $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. Let the horizontal distance from the airplane to ship 1 be $d_1$. Then $\tan72^{\circ}=\frac{34770}{d_1}$, so $d_1=\frac{34770}{\tan72^{\circ}}\approx\frac{34770}{3.07768}\approx11300$ feet. In meters, $d_1\div3.28084\approx11300\div3.28084\approx3444$ meters.
Step2: Convert units and set up for ship 2
The altitude of the airplane is still considered from the ocean - level. The height of cruise - ship 2 is $76$ meters, which is $76\times3.28084 = 249.34384\approx249$ feet. The airplane is at point B, $3000$ feet from point A. The vertical distance from the airplane to the top of ship 2 is $h_2 = 35000-249 = 34751$ feet. The angle of depression to the top of ship 2 is $\theta_2=54^{\circ}$. Let the horizontal distance from the airplane to ship 2 be $d_2$. Then $\tan54^{\circ}=\frac{34751}{d_2}$, so $d_2=\frac{34751}{\tan54^{\circ}}\approx\frac{34751}{1.37638}\approx25248$ feet. In meters, $d_2\div3.28084\approx25248\div3.28084\approx7708$ meters.
Step3: Find horizontal distance between yacht and ship 1
The airplane descends $2500$ meters ($2500\times3.28084 = 8202.1$ feet) when above ship 1. So the new altitude of the airplane above the ocean - level is $35000 - 8202.1=26797.9$ feet. The height of the yacht is $328$ feet. The vertical distance from the airplane to the top of the yacht is $h_3=26797.9 - 328=26469.9$ feet. The angle of depression to the top of the yacht is $\theta_3 = 32^{\circ}$. Let the horizontal distance from the airplane to the yacht be $d_3$. Then $\tan32^{\circ}=\frac{26469.9}{d_3}$, so $d_3=\frac{26469.9}{\tan32^{\circ}}\approx\frac{26469.9}{0.62487}\approx42360$ feet. The horizontal distance from the yacht to ship 1 is the difference in their horizontal distances from the airplane (assuming they are in the same horizontal - plane projection). First, we need to find the horizontal distance from the airplane to ship 1 in the new altitude situation. The vertical distance from the new - altitude airplane to ship 1 (height of ship 1 in feet is 230) is $26797.9 - 230=26567.9$ feet. Let this horizontal distance be $d_{1n}$. Then $\tan72^{\circ}=\frac{26567.9}{d_{1n}}$, so $d_{1n}=\frac{26567.9}{\tan72^{\circ}}\approx\frac{26567.9}{3.07768}\approx8632$ feet. The difference in horizontal distances $d_{y - s1}=42360 - 8632=33728$ feet, in meters $33728\div3.28084\approx10280$ meters.
Step4: Find horizontal distance between lighthouse and ship 2
The height of the lighthouse is $63$ meters, which is $63\times3.28084 = 206.69292\approx207$ feet. The vertical distance from the airplane (at the altitude when seeing the lighthouse) to the top of the lighthouse is $h_4=26797.9 - 207=26590.9$ feet. The angle of depression to the top of the lighthouse is $\theta_4 = 13^{\circ}$. Let the horizontal distance from the airplane to the lighthouse be $d_4$. Then $\tan13^{\circ}=\frac{26590.9}{d_4}$, so $d_4=\frac{26590.9}{\tan13^{\circ}}\approx\frac{26590.9}{0.23087}\approx115177$ feet. We know the horizontal distance from the airplane to ship 2 is $…
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a) (Diagram description: Draw a horizontal line to represent the ocean - level. Mark points A and B on a horizontal line above the ocean - level to represent the airplane's positions. Draw vertical lines down from A and B to represent the altitudes of the airplane. Mark the positions of cruise - ship 1, cruise - ship 2, yacht and lighthouse on the ocean - level or slightly above (accounting for their heights). Label the altitudes, horizontal distances, angles of depression and heights of the objects as per the problem description.)
b) Horizontal distance to ship 1: 11300 feet, 3444 meters; Horizontal distance to ship 2: 25248 feet, 7708 meters
c) 10280 meters
d) 27410 meters
e) 35105 meters