Answer:
a)(2.82, 1.03) and (-2.68, 2.25)
b) 5.63 m
Explanation:
a) The Cartesian coordinates of point (3.00 m, 20.0°) is
(3cos20.0°, 3sin20.0°) = (2.82, 1.03) m
The Cartesian coordinates of point (3.5 m, 140.0°) is
(3.5cos140.0°, 3.5sin140.0°) = (-2.68, 2.25) m
b) The distance between them
[tex]d = \sqrt{(2.82 - (-2.68))^2 + (1.03 - 2.25)^2} = \sqrt{5.5^2 + 1.22^2} = 5.63m[/tex]
(a) Cartesian coordinate (2.8191m, 1.0260m) and (-2.6810m, 2.2498m)
(b) distance = 5.6346m
Given a polar coordinate (r, θ);
Its Cartesian coordinate is (x, y) where;
x = r cos θ
y = r sin θ
Now, given the polar coordinate (3.00, 20.0⁰)
Its Cartesian coordinate is (x, y) where
x = 3.00 cos 20.0° = 3.00 x 0.9397 = 2.8191m
y = 3.00 sin 20.0° = 3.00 x 0.3420 = 1.0260m
Therefore,
Polar coordinate (3.00m, 20.0°) = Cartesian coordinate (2.8191m, 1.0260m)
Also,
Now, given the polar coordinate (3.50, 140.0⁰)
Its Cartesian coordinate is (x, y) where
x = 3.50 cos 140.0° = 3.50 x -0.7660 = -2.6810m
y = 3.40 sin 140.0° = 3.50 x 0.6428 = 2.2498m
Therefore,
Polar coordinate (3.50m, 140.0°) = Cartesian coordinate (-2.6810m, 2.2498m)
(a) Polar coordinates (3.00m, 20.0°) and (3.50m, 140.0°) = Cartesian coordinate (2.8191m, 1.0260m) and (-2.6810m, 2.2498m)
(b) To calculate the distance between them, it is easier to use the values from the Cartesian representation as follows;
distance = [tex]\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} }[/tex]
where;
([tex]x_{1}[/tex], [tex]y_{1}[/tex]) = (2.8191m, 1.0260m) and
([tex]x_{2}[/tex], [tex]y_{2}[/tex]) = (-2.6810m, 2.2498m)
=> distance = [tex]\sqrt{(-2.6810 - 2.8191)^{2} + (2.2498 - 1.0260)^{2} }[/tex]
=> distance = [tex]\sqrt{(-5.5001)^{2} + (1.2238)^{2} }[/tex]
=> distance = [tex]\sqrt{30.2511+1.4977}[/tex]
=> distance = [tex]\sqrt{31.7488}[/tex]
=> distance = 5.6346m
