Law of Sines and Cosines- Practice for Wednesday’s Quiz

1. By applying the Law of Sines and the Law of Cosines, try to figure out the total area of the figure below.



  1. A mountain climber is in the base of two mountains. He is trying to decide how much rope is needed to get to the top of the tallest mountain. He knows that from the top of the smaller mountain to the top of the taller mountain that he wants to climb is 100 meters apart and that the smaller mountain is 85 meters tall. He also knows that there is a 37 angle between both mountains. Figure out how much rope is needed to get to the top.
  1. A farmer owns a plot of land, but has decided that he wants to move to a different town. He wants to sell his land, but part of the paper that had the size of his land on it had been burned up. The piece that remained had the following information: Side a = 300 meters, Side b = 400meters Side c = 600 meters, and Side d = 500 meters. The angle between side b and side c was 76. The rest of the information was missing. Using this information and the diagram below, help the farmer find the total area of his farmland.
  1. A family decides to go on a vacation and see the country. They start from their home and drive 200 miles in a straight direction until they come to a place where they would like to stay for a while. After they leave this place, they drive 100 more miles at an angle of 115, they again stay at another place. They finally decide they want to come home, but wonder if there is a shorter route, is there? Help them find a way home.
  1. There is a shipwreck at sea and the ship sends out a signal for rescue to two separate stations. Station A is located 80 miles from station B, and the distress signal was picked up by station A at 43 from the ship, and by station B at 61 from the ship. Find out how far each station is from the ship and which one should send a rescue ship.

Law of Sines and Cosines.

Appendix.

Answers for Practice Questions 1-10.

1. a = b

SinA SinB

27 = 35

SinA Sin48

35SinA = 27Sin48

SinA = 27Sin48

35

A = Sin-1(0.5733)

A = 35

C = 180 – 48 – 35 = 97

c2 = a2 + b2 – 2ab CosC

c2 = 27 + 35 – 2 (27)(35) Cos97

c2 = 729 + 1225 – 2 (945) Cos97

c2 = 1954 – 1890 Cos97

c2 = 1954 +230.33

c2 = 2184.33

c = 46.7

2. a2 = b2 + c2 – 2bc CosA

32 = 52 + 72 – 2 (5)(7) CosA

9 = 25 + 49 – 2 (35) CosA

9 = 74 – 70 CosA

-65 = -70 CosA

0.9286 = CosA

Cos-1 0.9286 = A

A = 22

3.c2 = a2 + b2 – 2ab CosC

c2 = 112 + 152 – 2 (11)(15) Cos88

c2 = 121 + 225 – 2 (165) Cos88

c2 = 346 – 330 Cos88

c2 = 346 + 11.517

c2 = 357.517

c = 18.9

4. a = b

SinA SinB

45 = b

Sin48 Sin62

bSin48 = 45Sin62

b = 45Sin62

Sin48

b = 53.5

C = 180 – 48 – 62 = 70

c2 = a2 + b2 – 2ab CosC

c2 = 452 + 53.52 – 2 (45)(53.5) Cos70

c2 = 2025 + 2862.25 – 2 (2407.5) Cos70

c2 = 4887.25 – 4815 Cos70

c2 = 4887.25 + 1646.827

c2 = 6534.077

c = 80.8

5. b = c

SinB SinC

6.5 = 8.5

Sin41 SinC

6.5SinC = 8.5Sin41

SinC = 8.5Sin41

6.5

C = Sin-1 (0.8579)

C = 59

A = 180 – 41 – 59 = 80

a2 = b2 + c2 – 2bc CosA

a2 = 6.52 + 8.52 – 2 (6.5)(8.5) Cos80

a2 = 42.25 + 72.25 – 2 (55.25) Cos80

a2 = 114.5 – 110.5 Cos80

a2 = 114.5 - 19.188

a2 = 95.312

a = 9.8

6. A = 180 – 41 – 51 = 88

a = c

SinA SinC

a = 100

Sin88 Sin51

aSin51 = 100Sin88

a = 100Sin88

Sin51

a = 128.6

7. b = c

SinB SinC

b = 30

Sin110 Sin45

bSin45 = 30Sin110

b = 30Sin110

Sin45

b = 40

8. a = b

SinA SinB

8 = 5

Sin35 SinB

8SinB = 5Sin35

SinB = 5Sin35

8

B = Sin-1 0.3585

B = 21

C = 180 – 35 – 21 = 124

c2 = a2 + b2 – 2ab CosC

c2 = 8 + 5 – 2 (8)(5) Cos124

c2 = 64 + 25 – 2 (40) Cos124

c2 = 89 – 80 Cos124

c2 = 89 + 44.74

c2 = 133.74

c = 11.6

9.b2 = a2 + c2 – 2ab CosB

b2 = 22.1 + 17.5 – 2 (22.1)(17.5) Cos109

b2 = 488.41 + 306.25 – 2 (386.75) Cos109

b2 = 794.66 – 773.5 Cos109

b2 = 794.66 + 251.83

b2 = 1046.49

b = 32.3

a = b

SinA SinB

22.1 = 32.3

SinA Sin109

32.3SinA = 22.1Sin109

SinA = 22.1Sin109

32.3

A = Sin-10.6469

A = 40

C = 180 – 40 – 109 = 31

10. c2 = a2 + b2 – 2ab CosC

c2 = 302 + 522 – 2 (30)(52) Cos65

c2 = 900 + 2704 – 2 (1560) Cos65

c2 = 3604 – 3120 Cos65

c2 = 3604 – 1318.57

c2 = 4922.57

c = 70.2

c2 = a2 + b2 – 2ab CosC

70.22 = 382 + 422 – 2 (38)(42) CosC

4928.04 = 1444 + 1764 – 2 (1596) CosC

4928.04 = 3208 - 3192 CosC

1720.04 = – 3192 CosC

-0.5389 = CosC

Cos-1 – 0.5389 = C

C = 123

Area 1 = ½ ab SinC

Area 1 = ½ (30)(52) Sin65

Area 1 = ½ 1560 Sin65

Area 1 =706.922

Area 2 = ½ ab SinC

Area 2 = ½ (38)(42) Sin123

Area 2 = ½ 1596 Sin123

Area 2 = 669.262

Total Area = Area 1 + Area 2

Total Area = 706.922 + 669.262

Total Area = 1376.22

Answers to the Quiz

1. a = b

SinA SinB

85 = 100

SinA Sin37

100SinA = 85Sin37

SinA = 85Sin37

100

A = Sin 0.5115

A = 31

C = 180 – 31 – 37 = 112

c2 = a2 + b2 – 2 ab CosC

c2 = 852 + 1002 – 2 (85)(100) Cos112

c2 = 7225 + 10000 – 2 (8500) Cos112

c2 = 17225 – 17000 Cos112

c2 = 17225 + 6368.3

c2 = 23593.3

c = 153.6 meters

2. x2 = b2 + c2 – 2 ab CosC

x2 = 4002 + 6002 – 2 (400)(600) Cos76

x2 = 160000 + 360000 – 2 (240000) Cos76

x2 = 520000 – 480000 Cos76

x2 = 520000 – 116122.51

x2 = 403877.5

x = 635.5 meters

c2 = a2 + b2 – 2 ab CosC

635.52 = 3002 + 5002 – 2 (300)(500) Cos

403860.25 = 90000 + 250000 – 2 (150000) CosC

403860.25 = 340000 - 300000 CosC

63860.25 = -300000CosC

-0.213 = CosC

C = Cos-1-0.213

C =102

Total Area = Area 1 + Area 2

Total Area = ½ ab SinC + ½ ab Sin C

Total Area = ½ (400)(600) Sin76 + ½ (300)(500) Sin102

Total Area = ½ 240000Sin76 + ½ 150000Sin102

Total Area = ½ 232871 + ½ 146722

Total Area = 116436 + 73361

Total Area = 1897972 meters

3. c2 = 1002 + 2002 – 2 (100)(200) Cos115

c2 = 10000 + 40000 – 2 (20000) Cos115

c2 = 50000 – 40000 Cos115

c2 = 17225 + 16905

c2 = 34130

c = 185 miles

Yes, there is a much shorter route home if they could drive straight back home from where they were it would only be 185 miles compared to 300 miles.

4.C = 180 – 61 – 43 = 76

a = c

SinA SinC

a = 80

Sin43 Sin76

aSin76 = 80Sin43

a = 80Sin43

Sin76

a = 56 miles

a = b

SinA SinB

56 = b

Sin43 Sin61

56Sin61 = bSin43

b = 56Sin61

Sin43

b = 72 miles

It would make more sense for Station A to send out a rescue vessel because they are only 56 miles from the shipwreck so it would be closer to save them.