An angle's initial ray points in the 12-o'clock direction and the terminal ray rotates CCW. Let θ represent angle's varying measure (in radians).
a. If θ=0.5 what is the slope of the terminal ray?
b. Write an expression (in terms of θ) that represents the varying slope of the terminal ray.

The slope of the rotating terminal ray with the initial ray at 12-O'clock is

given by the cotangent of the angle of rotation.

Reasons:

The direction of the initial ray =  12-O'clock direction

The direction of rotation of the terminal ray = CCW

θ = The measure of the angle formed by the terminal ray

a. Required:

The value of the slope of the terminal ray when θ = 0.5

Solution;

The slope is the ratio of the rise to the run of the terminal ray, which is

given by the tangent of the angle made by the terminal ray with the

horizontal, as follows;

The angle the terminal ray makes with the horizontal =  

b. The slope is the tangent the terminal ray makes with the horizontal,

therefore, the expression in terms of θ that represents the varying slope is

presented as follows;

Learn more here:

link

Step-by-step explanation:

Given that, the initial position of the clock is 12°Clock

The clock rotates counter clockwise

Given that θ is in radian angle and we know that

2π radian = 360°

Then, 1 radian = 360 / 2π

1 radian = 57.3°

A. Given that the clock rotates θ=0.5rad, so, the clock rotates at an angle of

θ=0.5 × 57.3 = 28.65°

To, from the x-axis, the angle is

90+28.65 = 118.65°

Then, the slope can be determine using

m = tanθ

m = tan118.65

m = -1.83

B. Write an expression (in terms of θ) that represents the varying slope of the terminal ray.

The expression of the slope that is varying with θ is given as

m = tanθ

Where m is slope and θ is the rotation of the clock or the angle of rotation

b, i believe, but not sure.

need more info

step-by-step explanation: