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Warm-up. A ship leaves port at noon and heads due west at 20 knots (nautical miles per hour). At 2 pm, the ship changes course to N54 ˚W. Find the ship’s bearing and distance from the port of departure at 3 pm. Draw and label a sketch of the problem. We will solve it together.
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Warm-up A ship leaves port at noon and heads due west at 20 knots (nautical miles per hour). At 2 pm, the ship changes course to N54˚W. Find the ship’s bearing and distance from the port of departure at 3 pm. Draw and label a sketch of the problem. We will solve it together. 4.8 Applications and Models
SOLUTION: A ship leaves port at noon and heads due west at 20 knots (nautical miles per hour). At 2 pm the ship changes course to N54W. Find the ship’s bearing and distance from the port of departure at 3 pm (40 + 16.18)2 + (11.75)2 = d2 d = 57.39nm d 20 nm x 54 36 y 40nm z = 2(20 nm)
A ship leaves port at noon and heads due west at 20 knots (nautical miles per hour). At 2 pm the ship changes course to N54W. Find the ship’s bearing and distance from the port of departure at 3 pm z = 12 W 12 N or N 78 W d 20 nm x 54 36 y 40nm z = 2(20 nm)
Example 4 SOLVING A PROBLEM INVOLVING ANGLES OF ELEVATION From a given point on the ground, the angle of elevation to the top of a tree is 36.7˚. From a second point, 50 feet back, the angle of elevation to the top of the tree is 22.2˚. Find the height of the tree to the nearest foot. The figure shows two unknowns: x and h. Since nothing is given about the length of the hypotenuse, of either triangle, use a ratio that does not involve the hypotenuse. TANGENT
Example 4 In triangle BCD: SOLVING A PROBLEM INVOLVING ANGLES OF ELEVATION (continued) In triangle ABC: Each expression equals h, so the expressions must be equal.
Example 4 SOLVING A PROBLEM INVOLVING ANGLES OF ELEVATION (continued) Since h = x tan 36.7˚, we can substitute. The tree is about 45 feet tall.
Example: Solving a Problem Involving Angles of Elevation Sean wants to know the height of a Ferris wheel. From a given point on the ground, he finds the angle of elevation to the top of the Ferris wheel is 42.3˚. He then moves back 75 ft. From the second point, the angle of elevation to the top of the Ferris wheel is 25.4˚. Find the height of the Ferris wheel.
The figure shows two unknowns: x and h. Since nothing is given about the length of the hypotenuse, of either triangle, use a ratio that does not involve the hypotenuse, (the tangent ratio). In triangle ABC, In triangle BCD, B h C x A D 75 ft
Since each expression equals h, the expressions must be equal to each other. Solve for x. Distributive Property Get x-terms on one side. Factor out x. Divide by the coefficient of x.
We saw above that Substituting for x. • tan 42.3 = .9099299 and tan 25.4 = .4748349. So, tan 42.3 - tan 25.4 = .9099299 - .4748349 = .435095 and • The height of the Ferris wheel is approximately 75 ft.
Example: Solving a Problem Involving Angles of Elevation • Jenn wants to know the height of a Ferris wheel. From a given point on the ground, she finds the angle of elevation to the top of the Ferris wheel is 49.6° . She then moves back 65 ft. From the second point, the angle of elevation to the top of the Ferris wheel is 29.7° . Find the height of the Ferris wheel.
The figure shows two unknowns: x and h. Since nothing is given about the length of the hypotenuse, of either triangle, use a ratio that does not involve the hypotenuse, (the tangent ratio). In triangle ABC, In triangle BCD, B h C x A D 75 ft x
Since each expression equals h, the expressions must be equal to each other. Solve for x. Distributive Property Get x-terms on one side. Factor out x. Divide by the coefficient of x.
B h C x A D 75 ft x≈61.32 ft x
Marla needs to find the height of a building. From a given point on the ground, she finds that the angle of elevation to the top of the building is 74.2°. She then walks back 35 feet. From the second point, the angle of elevation to the top of the building is 51.8°. Find the height of the building.
In triangle ABC In triangle BCD Solving a Problem Involving Angles of Elevation (cont.) • There are two unknowns, the distance from the base of the building, x, and the height of the building, h.
Set the two expressions for h equal and solve for x. Since h = x tan 74.2°, substitute the expression for x to find h. The building is about 69 feet tall.