# Vectors Math Problem Help?

a) Determine the magnitude of the resultant vector, u + v, to the nearest tenth of a newton.

b) Determine the direction of the resultant vector, u + v, to the nearest degree.

### 7 Answers

- TomVLv 710 months agoFavorite Answer
Resultant Vector:

<R> = <30 + 50cos42°, 50sin42°>

a) |R| = √[(30 + 50cos42°)² + (50sin42°)²]

= √(900 + 3000cos42° + 2500)

= √(3400 + 3000cos42°)

≈ 75.0 N (to the nearest 10th)

b) Arg<R> = arctan[50sin42°/(30 + 50cos42°)]

≈ arctan(33.45653/67.15724)

≈ 26° (to the nearest degree)

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- oldschoolLv 710 months ago
Sum the horizontal components:

30+50*cos42 = 67.2

Sum the vertical components:

50*sin42 = 33.5

arctan 33.5/67.2 = 26.5° <<<<<

67.2²+33.5² = 75²

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- Mike GLv 710 months ago
s = resultant vector magnitude

θ = direction

Cosine Rule

s^2 = 30^2+50^2-2*30*50cos138

s = 75 N

sinθ/50 = sin138/75.03

θ = 26.5°

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- Ian HLv 710 months ago
M^2 = x^2 + y^2 = (30 + 50c)^2 + (50s)^2

M^2 = 2500(s^2 + c^2) + 900 + 3000c = 3400 + 3000cos42

M ~ √[5629.43] ~ 72. 0296 or 72.0 to nearest tenth

tan(θ) = 50 sin(42)/[30 + 50 cos(42)]

θ = artan(0.498182) ~ 26.48 degrees or 26 to nearest degree.

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- ted sLv 710 months ago
U + V = < - 30 + 50 cos 42° , 50 sin 42° >...| U + V | = √ ( 2590 - 600 cos 42° )

tan Θ = [ 50 sin 42° / (- 30 +50 cos 42°) ]....Θ ≈102 °

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- RealProLv 710 months ago
It's already drawn for you...?

Magnitude sqrt[ (50sin42°)^2 + (30+50cos42°)^2 ]

Direction arctan[ 50sin42° / (30+50cos42°) ] counterclockwise from horizontal

- ted sLv 710 months agoReport
should be ' - 30 '

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- davidLv 710 months ago
180 - 42 = 138

c^2 = 30^2 + 50^2 - 2*30*50*cos138 = 5629.43

c = 75.03 N <<< mag. or resultant

sin X / 50 = sin 138 / 75.03

x = 26.48 degrees <<< angle for the resultant

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