Question

Two points in a plane have polar coordinates (2.50 m, 30.0°) and (3.80 m, 120.0°). Determine (a) the Cartesian coordinates of these points and (b) the distance between them.

Answer 1

`x_1 = 2.16 m`

`y_1 = 1.25 m`

`x_2 = -1.90`

`y_2 = 3.29 m`

Part b)

`d = 4.54 m`

Step-by-step explanation:

Part a)

Polar coordinates is given as

`r_1 = (2.50 m, 30^o)`

so we have

`x = r cos\theta`

`y = r sin\theta`

`x_1 = 2.50 cos30 = 2.16 m`

`y_1 = 2.50 sin30 = 1.25 m`

similarly we have

`r_2 = (3.80, 120 ^o)`

`x_2 = 3.80 cos120 = -1.90`

`y_2 = 3.80 sin120 = 3.29 m`

Part b)

now the distance between two coordinates

`d = √((x_2 - x_1)^2 + (y_2 - y_1)^2)`

`d = √((-1.90 - 2.16)^2 + (3.29 - 1.25)^2)`

`d = 4.54 m`

Answer 2

a) Cartesian coordinates of (2.50 m, 30.0°) = (2.17.1.25)

Cartesian coordinates of (3.80 m, 120.0°) = (-1.90.3.29)

b) Distance between (2.17.1.25) and (-1.90.3.29) = 4.55 meter.

Step-by-step explanation:

Points in polar coordinates = (2.50 m, 30.0°) and (3.80 m, 120.0°)

(2.50 m, 30.0°) = (2.50*cos 30, 2.50*sin 30) = (2.17.1.25)

(3.80 m, 120.0°) = (3.80*cos 120, 3.80*sin 120) = (-1.90.3.29)

a) Cartesian coordinates of (2.50 m, 30.0°) = (2.17.1.25)

Cartesian coordinates of (3.80 m, 120.0°) = (-1.90.3.29)

b) We have distance between (a,b) and (c,d) by distance formula
`√((c-a)^2+(d-b)^2)`

So distance between (2.17.1.25) and (-1.90.3.29) =
`√((-1.9-2.17)^2+(3.29-1.25)^2)=√(20.7265)=4.55 meter`

Distance between (2.17.1.25) and (-1.90.3.29) = 4.55 meter.