# Determine two pairs of polar coordinates for the point (3, 3) with 0° ≤ θ < 360°.

**For Solution:**

The two pairs of polar coordinates can be determined by understanding the diagram below:

The diagram above is a comprehensive representation of both polar coordinates and cartesian coordinates. Point A has the given coordinates of (3, 3).

From these coordinates we can determine the angle θ (∠BOA) as follows:

We know,

x = rcosθ --- (1)

y = rsinθ --- (2)

Dividing (2) by (1) we get

y/x = tanθ

x = 3 and y = 3 (given coordinates of point A) we have

Tanθ = 3/3 = 1 which implies

θ = 45° or π/4 radians

Also the radius can be determined from the right-angled triangle △AOB with the help of pythagorean theorem:

OA^{2 }= r^{2} = OB^{2} + BA^{2} = 3^{2} + 3^{2}

OA^{2} = r^{2} = 9 + 9 =18

OA = r = √18 = 3√2

Polar Coordinate is given by (r, θ), and therefore the polar coordinate of point A is (3√2, π/4).

We can also state that the ray OA has the polar coordinate (3√2, π/4) and the corresponding cartesian coordinate is (3, 3).

We can get the second polar coordinate by moving to the other side of the pole (to the second quadrant) another point (-3, 3) which is at an angle of 180 - π/4 from the original ray OB. The ray r = 3√2 moves to the second quadrant and becomes OA’. The angle it ∠A’OB is 180° - 45° or π - π/4 = 3π/4. Since the ray OA’ is now on the other side of the pole (y-axis) the direction of the ray becomes opposite and hence r = -3√2

Hence the second polar coordinate becomes (-3√2, 3π/4).

The cartesian coordinates for the point A’ are then as follows:

x = rcosθ = 3√2cos(3π/4) = 3√2(-1/√2) = -3

y = rsinθ = 3√2sin(3π/4) = 3√2(1/√2) = 3

The polar coordinate (-3√2, 3π/4) thus has the cartesian coordinate of (-3,3) which is the point A’.

The next pair of Polar coordinate of (3,3) is point A’’ which is obtained by rotating counterclockwise with a rotation of π + π/4 =+5π/4 to reach point A’’. The radius r is 3√2 and since it is on the left-hand side of the pole (y-axis) it is given a negative sign.

Therefore the polar coordinates of the A’’ is (-3√2, 5π/4).

The cartesian coordinate for point A’’ is given by

x = rcosθ = 3√2cos(5π/4) = 3√2(-1/√2) = -3

y = rsinθ = 3√2sin(5π/4) = 3√2(1/√2) = -3

So the cartesian coordinates for point A’’ is given by (-3,-3).

The fourth polar coordinate is generated by rotating counterclockwise 3π/2 + π/4 = 7π/4 to obtain point A’’’.

At this point, r will be 3√2 and the sign attached to r will be +ve. So the polar coordinate for point A’’’ will be (3√2, 7π/4).

x = rcosθ = 3√2cos(7π/4) or 3√2cos(-π/4 )= 3√2(1/√2) = 3

y = rsinθ = 3√2sin(7π/4) or or 3√2sin(-π/4) = 3√2(1/√2) = -3

So cartesian points for point A’’’ are (3, -3)

## Determine two pairs of polar coordinates for the point (3, 3) with 0° ≤ θ < 360°.

**Summary:**

The two pairs of polar coordinates for the cartesian coordinate between 0° and 360° are (3√2, π/4) & (-3√2, 3π/4), (-3√2, 5π/4) & (3√2, 7π/4).

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