what are the horizontal and vertical components of a vector that is 25units long with an angle of 130 degrees​

The horizontal component of the vector is approximately -16.07

The vertical component of the vector is approximately 19.15

Explanation:

A vector quantity is a quantity that posses both magnitude and direction specifications

The magnitude of the given vector, R = 25 units

The direction of the vector, θ = 130°

Therefore the location of the vector, starting from the origin of the Cartesian plane, is in the second quadrant having an angle of 180° - 130° = 50° to the horizontal

Given that the x-coordinate  is negative in the second quadrant, we have;

The horizontal component of the vector, Rₓ = The x-coordinate of the vector = R×cos(θ)

∴ The horizontal component of the vector, Rₓ = 25 × cos(130°) = -25 × cos(50°) ≈ -16.07

The horizontal component of the vector, Rₓ ≈ -16.07

The vertical component of the vector, = The y-coordinate of the vector = R×sin(θ)

∴ The vertical component of the vector, = 25 × sin(130°) ≈ 19.15

The vertical component of the vector, ≈ 19.15

The vector can be resolved as R = -16.07·i + 19.15·j

your answer is c! hope this : )

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west