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Algebraic Vectors A An X, Thesis

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66km at 30.25 degrees above the horizon C

b

Displacement is 49m at 7.43 degrees East of North (assuming North for downfield)

8m

3.5m

5m

C

A

B

a c b

D

C = 90 E = 180 -- D = 35

D = 90 + 55 = 145 F = 180 -- E -- C = 55

3.5/sin90 = f/sin55 3.5/sin90 = e/sin35

3.5 x sin55 = f 3.5 x sin35 = e f = 2.87 e = 2.01

b = f + 8 = 10.87 a = e + 5 = 7.01

c2 = 10.872 + 7.012

c2 = 167.297

c = 12.93 = magnitude

12.93/sin90 = 7.01/sinA

12.93 x sinA = 7.01

sinA = 7.01/12.93 = .54

A = 32.8 = direction

Displacement is 12.93 meters 32.8 degrees East of North

E

F

f e

60 km

c

C

A

B

D = 30 E = 60

0
F = 90 -- D = 60 C = 180 -- E -- G

G = 180 -- 90 -- F = 30 C = 180 -- 60 -- 30 = 90

c2 = 602 + 1552

c2 = 27625

c = 166.201 = magnitude

166.201/sin90 = 155/sinA

166.201 x sinA = 155

sinA = 155/166.201 = .933

A = 68.84 = direction

Displacement is 166.201 km at 68.84 degrees West of North

D

E

F

G

5m/s

3m/s c2 = 32 + 52 5.83/sin90 = 3/sinA

c2 = 9 + 25 = 34 5.83 x sinA = 3

c = 5.83 = magnitude sinA = 3/5.83

sinA = .514

A = 30.97 = direction

Displacement is 5.83 meters/second at 30.97 degrees downstream of perpendicularity with the flow of the river

A

c

1m/s

6m/s c

A

c2 = 12 + 62 6.08/sin90 = 6/sinA

c2 = 1 + 36 = 37 6.08 x sinA…

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