Question

Let vector B = 5.50 m at 60°. Let vector C have the same magnitude as vector A and a direction angle greater than that of vector A by 25°. Let vector A · vector B = 27.0 m2 and vector B · vector C = 34.5 m2. Find the magnitude and direction of vector A.

Answer 1

write the vector A as (Acosθ, Asinθ)
B as (2.75, 4.76)
C as (Acos(θ+25), Asin(θ+25))

A.B = 27,

so Acosθ * 2.75 + Asinθ * 4.76 = 27 ................... (1)

B.C = 34.5,

so Acos(θ+25) * 2.75 + Asin(θ+25) * 4.76 = 34.5......(2)

(1) and (2) are two equations in two unknowns A and θ, so we can solve them fairly easily.

start off with simplifying the LHS of (2)

cos (θ+25) = cosθ*cos25 - sinθ*sin25 = 0.906cosθ - 0.423sinθ
sin(θ+25) = cosθ*sin25 + sinθ*cos25 = 0.423cosθ + 0.906sinθ

A*(0.906cosθ - 0.423sinθ) * 2.75 + A*(0.423cosθ + 0.906sinθ) * 4.26 = 34.5 ... which gives

2.6963 A sinθ + 4.29 A cosθ = 34.5 .........(3)

rewrite (1):
Acosθ * 2.75 + Asinθ * 4.76 = 27

Now you can eliminate A from either equation, and use sin² + cos² = 1 to get all sin's or all cos's, and take the inverse etc. It's long and messy, but nothing difficult.

approximately A = 7.21, θ = 12.88 degrees at the end of it all.