The cartesian coordinates of a point are given. (a) (6, −6) (i) find polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π. (r, θ) = $$ correct: your answer is correct. (ii) find polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π. (r, θ) = $$ correct: your answer is correct. (b) (−1, 3 ) (i) find polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π. (r, θ) = $$ correct: your answer is correct. (ii) find polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π. (r, θ) = $$ correct: your answer is correct.

Part a)

(i) The polar coordinates are

(ii) The polar coordinates are

Part b)

(i)The polar coordinates are

(ii)The polar coordinates are

Step-by-step explanation:

Part a) we have

(6,-6)

(i) Find polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π

step 1

Find r

substitute

step 2

Find angle theta

substitute

Remember that the point (6,-6) lies on IV quadrant

so

The angle theta is equal to

therefore

The polar coordinates are

(ii) Find polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π.

For r< 0 the point lies on II Quadrant

so

the angle theta is equal to

therefore

The polar coordinates are

Part b) we have

(-1,3)

(i) Find polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π

step 1

Find r

substitute

step 2

Find angle theta

substitute

Remember that the point (-1,3) lies on II quadrant

so

The angle theta is equal to

therefore

The polar coordinates are

(ii) Find polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π.

For r< 0 the point lies on IV Quadrant

so

the angle theta is equal to

therefore

The polar coordinates are

u bout ugly as   a mf

step-by-step explanation: lol