More calculus! (i'm so sorry)

1. given the curve that is described by the equation r = 3 cos ф, find the angle that the tangent line makes with the radius vector when ф = 120°
a. 30°, b. 45°, c. 60°, d. 90°

2. a line rotates in a horizontal plane according to the equation ф = 2t^3 - 6t, where ф is the angular position of the rotating line, in radians, and t is the time, in seconds. determine the angular acceleration when t = 2 sec.
a. 6 radians per second^2, b. 12 radians per second^2, c. 18 radians per second^2, d. 24 radians per second^2

to get brainliest you have to be able to give me at least some reasoning or proof for the answers so that i know you didn't just get them off of google and they're legit answers.

Recall that converting from Cartesian to polar coordinates involves the identities



As a function in polar coordinates, depends on , so you can write .

Differentiating the identities with respect to gives



The slope of the tangent line to is given by



Given , you have . So the tangent line to has a slope of



When , the tangent line has slope



This line is tangent to the point which in Cartesian coordinates is equivalent to , so the equation of the tangent line is



In polar coordinates, this line has equation




The tangent line passes through the y-axis when , so the y-intercept is .

The vector from this point to the point of tangency on is given by the difference of the vector from the origin to the y-intercept (which I'll denote ) and the vector from the origin to the point of tangency (denoted by ). In the attached graphic, this corresponds to the green arrow.



The angle between this vector and the vector pointing to the point of tangency is what you're looking for. This is given by





The second problem is just a matter of computing the second derivative of with respect to and plugging in .