Law of Sines and Cosines



Homework: Law of Sines and Cosines Worksheet #1-10

1. For the triangle below, where two sides and the angle opposite side b are known, solve for the remaining side and angles that are unknown. Side a = 27, side b = 35, and angle B = 48(.

2. For this Triangle All three sides are known, but Angle A is needed. Side a = 3, side b = 5, and side c = 7. Decide which method is best suited for this and solve for A.

3. In this case side a = 11, b = 15, and angle C = 88( are known, but you must find side c. Figure out which method you should use and apply it to find the length of side c.

4. Two angles and only one side of a triangle are known, Angle A = 48(, Angle B = 62(, and side a = 45. Find the missing angle and the two missing sides for the triangle.

5. Side b of this triangle has a length of 6.5, side c has a length of 8.5, and angle B has a degree measure of 41(. Find all missing information.

6. In a triangle where angle B = 41(, angle C = 51(, and side c = 100, Find the length of side a.

7. For a triangle with all measurements known except side b, use whichever formula you find easiest to solve for the missing side. Side a = 18, side c = 30, Angle A = 25(, Angle C = 45(, and angle B = 110(.

8. In a triangle where side a = 8, side b = 5, and Angle A = 35(, find the values for all of the missing sides or angles whichever the case may be by using bith the law of sines and the law of cosines.

9. A triangle has two known sides but one one known angle, which is between the two sides. Side a = 22.1, side c = 17.5, and Angle b = 109. Find all missing information.

10. By applying the Law of Sines and the Law of Cosines, try to figure out the total area of the figure below with the area formula being Area = ½ ab SinC.

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

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