Find all polar coordinate pairs of the point with Cartesian coordinates:
(-4sqrt(3), 4)
thanks for any help you can give me.
Here I drew a hand drawn picture.
The BIG point is the point you are given.
There are two angles the bigger and smaller (inside the triangle) which are shown as dotted lines.
The base of the triangle is the BIG black line.
The height of the triangle is the green line.
And the hypotenuse is the red line.
Now, we will find the smaller angles shown by the dotted angle.
Let theta be the angle.
Then we know that,
tan(theta)=OPPOSITE/ADJACENT=4/4sqrt(3)=1/sqrt(3)=sqrt(3)/3
Note it is not negative because we are finding the lengths, thus even though the x-value is negative we disregard it because the length is what we are trying to find. (In the last step I rationalized the denominator).
Okay so what angle gives you tangent of sqrt(3)/3. If you pay attention in class that is pi/6. Thus, the bigger dotted angle is what you need to find in order to convert to polar coordinates, that is pi-pi/6=5pi/6 because the full x-axis is pi radians.
Good now you know the oriented angle that you form. Meaning theta = 5pi/6. What is the radius. That is simple, look at the triangle. It is the sum of the squares of the sides by Pythagoren theorem. That is,
(4sqrt(3))^2+4^2=16*3+16=64
Now that square root, sqrt(64)=8.
Thus the radius is 8 units long.
Meaning r = 8.
Thus the polar coordinates are:
(5pi/6,8)
Hello, rcmango!
Find all polar coordinate pairs of the point with Cartesian coordinates: (-4sqrt(3), 4)
There are formulas for this problem:
Given: (x, y)
. . . . . . . . . . ______
Then: . r .= .√x² + y² . . . tan θ .= .y/x
We are given: .x = -4sqrt{3}, .y = 4
. . . . . . . . . . _______________ . . . .__
Then: . r .= .√(-4sqrt{3})² + 4² .= .√64 .= .8
. . . . . . . . . . . . . 4 . . . . . . .1
And: . tan θ .= . ------ . = . - ----
. . . . . . . . . . . .-4√3. . . . . .√3
. . Hence: .θ .= .5π/6, -π/6
We know that the angle is in Quadrant 2, so: .θ .= .5π/6
The complete solution is: .(8, 5π/6 + 2πk) for any integer k.