Practice Final, Part 2 – word problems

(1) The angle of elevation from a point 230 feet from the base of a building to the top of the building is 60°. Find the height of the building.

(2) An airplane traveling at 250 mph is descending at an angle of depression of 8 degrees. How many miles will the plane descend in 10 minutes?

(3) One ship leaves a port at a heading of 45º, travelling an average of 16 mph, at 2pm. Another ship leaves the same port at a heading of 105º, travelling an average of 23 mph, at 3pm. How far apart will the ships be at 5pm?

(4) (this is #36 from the text – there is a picture there). Two observers, in the same vertical plane as a kite and at a distance of 30 feet apart, observe the kite at angles of 62º and 78º. Find the height of the kite.

(5) A vacant lot is triangular in shape and priced at a value of $7.50/ft2. Find the price of the lot if it measures 100 by 106 by 110 feet.

(6) A satellite dish has the shape of a paraboloid. The signals that it receives are reflected to a receiver that is located at the focus of the paraboloid. If the dish is 6 feet across at its opening and 1 feet deep at its center, determine the location of its focus.

Solutions:

(1)

60° h

230

so

(2) First, find the horizontal distance traveled by the plane d = rt, or

. Then set up this super cool triangle:

42

8º

d

It appears that

(3) We have the following geometric representation:

a

46 48

60º

With the following explanations:

distance = rate * time:

46 = 23*(5pm – 3pm)

48 = 16*(5pm - 2pm)

60º =105º - 45º

We have SAS, so we can solve using the Law of Cosines:

(4) Kite

b

Obs1 Obs2

62º 78º

30

Note that one angle in the triangle will be the complement of the 78º angle, and will have measure 180-78 =102º, making the final angle 180 – (102+62) = 16º. We will then have ASA, and can solve for the height of the kite using the Law Of Sines:

(5) The cost of the lot will be area * cost/ft2. We can find the area using Heron's formula:

(6) As we discussed in class, the cross-section ('slice') of the dish laying below the receiver is a parabola, which is 6 feet wide. We could graph this cross-section as:

Because the graph points up, the square is on x and p > 0. The vertex (h, k) = (0,0), and the equation will have form: . We need to determine p – plug in (3,1) and solve for it.

is the distance from the vertex to the receiver/focus, that is 2.25 feet.