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# Can someone help me with trig please?

Change the rectangular coordinates to polar coordinates with r > 0 and 0 ≤ 𝜃 ≤ 2𝜋.

I'll attach the image

### 2 Answers

- L. E. GantLv 71 month ago
(a) r = sqrt(7^2 + (7sqrt(3))^2) = sqrt(49 + 49*3) = 14

theta = arctan(-sqrt(3)) = 2pi - pi/3 = 5pi/3

(look at the question -- you need RADIANS not degrees 5pi/3 = 300 degrees)

(b) r = 8sqrt(2) and pi/4 (pi/4 = 45 degrees)

- Daniel HLv 51 month ago
a)

h = sqrt(7^2 + (-7sqrt(3))^2) = 14

4th quad

arcsin(-7sqrt(3)/14) = 5pi/3

arccos(7/14) = 5pi/3

b)

h = sqrt(64 +64) = 8sqrt(2)

1st quad

arcsin(8/8sqrt(2)) = pi/4

arccos(8/8sqrt(2)) = pi/4

You were correct, but put degrees instead of rads