Question

I dont really know how to do this can somebody help?

Answer 1

9514 1404 393

Answer:

  8∠11π/6

Step-by-step explanation:

To simplify the way we talk of this number, we will refer to it as a vector. We'll assume its head is at the given coordinates, and its tail is at the origin. The magnitude of the vector is the root of the sum of the squares of its components:

  $\displaystyle r=\sqrt{x^2+y^2}=\sqrt{(4\sqrt{3})^2+(-4)^2}=\sqrt{48+16}=\sqrt{64}\\\\r=8$

The angle of the vector with respect to the +x axis is ...

  $\theta=\arctan{\dfrac{y}{x}}\qquad\text{a 4-quadrant arctangent}\\\\\theta=\arctan{\dfrac{-4}{4\sqrt{3}}}=-\arctan{\dfrac{1}{\sqrt{3}}}\qquad\text{a 4th-quadrant angle}\\\\\theta=\dfrac{11\pi}{6}$

In magnitude∠angle notation, the polar coordinates are ...

  (4√3, -4)  ⇔  8∠(11π/6) . . . . . θ in radians

_____

Additional comment

Spreadsheets generally have an ATAN2(x, y) function that will return a 4-quadrant version of the angle. The usual ATAN( ) function will give the angle only in quadrants I or IV.

Some graphing calculators may also have the ATAN2 function. Some will simply do the entire conversion for you, without any need to deal separately with magnitude and angle. (You may need to write the number as a complex number, x +iy.)

Answer 2

Answer:

(8, 11π/6)

Step-by-step explanation:

r= √(4√3)²+(-4)² = √48+16=√64=8

0 = tan~¹ -4/(4√3)

= tan~¹ -⅓√3 = 330° = 11π/6

the polar coord. => (8, 11π/6)