Law of Sines Word Problems (again) =[ I wouldn't ask if I knew how to set these up.?

Question

1. Two tracking stations, A and B, are on an east-west line 110 miles apart. A forest fire is located at F, on a bearing 42 degrees northeast of station A and 15 degrees northeast of station B. How far is the fire from station A?

2. On a ship sailing north, a woman notices that a hotel on a shore has a bearing of 20. A little while later, after having sailed 40km, she observes that the bearing of the hotel is now 100. How far is the ship from the hotel?

3. A ship is steaming south. The navigator notices that the bearing of a lighthouse is 120. After moving 8mph for 2 hours, he observes that the bearing of the lighthouse is 25. Find his distance from the lighthouse at the time of the 2nd sighting.

4. Bill determines that the angle of elevation to the top of the building measures 40 degrees, 30'. If he walks 102 ft. closer to the building, the measure of the new angle of elevation will be 50 degrees, 20'. Find the height of the building.

Answer 1

1.  northeast in this context has to mean E of N

Let

FAB = A = 90° - 42° = 48°

BF = a

ABF = B = 90° - 15° = 75°

AF = b

BFA = F = 180° - 48° - 75° = 57°

AB = f = 110 mi.

b/sinB = f/sinF

b = (110 mi.)(sin75°/sin57°

b ≈ 126.69 mi.

2.

A = 20°

B = 180° - 100° = 80°

C = 180° - 80° - 20° = 80°

c = 40 km

a = csinC/sinA

a = (40 km)(sin80°/sin20°)

3.

A = 180° - 120° = 60°

B = 25°

C = 180° - A - B

c = (8 mph)(2 h)

a = csinA/sinC

4. --different problem--

h = (x + 120)tan40.5°

h = xtan((50*60 + 20)/60)

OR (since you're practicing Law of Sines)

h = csinA

c = 102sin(180° - 50° 20')/sin(180° - 40.5° - (180° - 50° 20'))

h = 102sin(180° - 50° 20')sin(40.5°)/sin(180° - 40.5° - (180° - 50° 20'))