how do you find the angle of a vector when given both its sides?

Answer 1

Suppose the component vectors are a and b with magnitudes |a| and |b| and the angle between them θ

Then, on completing the parallelogram of forces, the angle between the end of a and the beginning of b is (180° - θ)

If the resultant vector is R (which is the required vector), makes an angle of α between a and R, then by the sine rule

sinα/|b| = sin(180° - θ)/|R|
So sinα = |b|sinθ/|R| (because sin(180° - θ) = sinθ
Hence α = arcsin{|b|sinθ/|R|}

Note If you need |R| then by the cosine rule:

|R|² = |a|² + |b|² - 2*|a|*|b|*cos(180° - θ)
= |a|² + |b|² + 2*|a|*|b|*cosθ (since cos(180° - θ) = - cosθ)

This is for a and b in any orientation on a plane (not necessarily at right angles)

IF θ = 90° then cosθ = 0 and sinθ = 1
and these equations reduce to

|R|² = |a|² + |b|²
and
α = arcsin{|b|/|R|} (which incidentally also equals arctan {|b|/|a|}

Answer 2

By "angle of a vector" I assume you mean the direction, and by "both its sides" I assume you mean the contributing vectors:

to solve, use the pythagorean theorem: a^2 + b^2 = c^2 where c would be the vector, and a and b would be the component vectors. Once you have a, b, and c, you can determine the angle/direction of the vector using trig functions:

sin = side (a or b) opposite the angle divided by the hypotenuse (c)

Best of luck!

Answer 3

Window shooper

Answer 4

Angle = acos((A.B)/(Mod(A)Mod(B)))