The coordinates of the point in the x^{prime} y^{prime} - coordinate system with the given angle of rotation and the xy-coordinates.

Question

The coordinates of the point in the $x^{'} y^{'}$ - coordinate system with the given angle of rotation and the xy-coordinates.

Answer 1

Suppose the x- and y- axes are rotated about the origin through a positive acute angle $θ$ , then the coordinates $(x, y)$ and $(x^{'}, y^{'})$ of a point P in the xy- and $x^{'}, y^{'}$ - coordinate systems are related by the following formulas:
$x^{'} = x \cos θ + y \sin θ$
$y^{'} = - x \sin θ + y \cos θ$
$x = x^{'} \cos θ - y^{'} \sin θ$
$y = x^{'} \sin θ + y^{'} \cos θ$
Given:
The angle of rotation is $θ = 30^{\circ}$ and the x- and y- coordinates are 0 and 2, respectively.
Calculation:
Use the definition, substitute the values of x-, y- coordinates and $θ$ in order to obtain the values of $x^{'}, y^{'}$ - coordinates.
$x^{'} = (0) \cos 30^{\circ} + (2) \sin 30^{\circ}$
$y^{'} = - (0) \sin 30^{\circ} + (2) \cos 30^{\circ}$
Know that $\sin 30^{\circ} = \frac{1}{2} and 30^{\circ} = \frac{\sqrt{3}}{2}$
Thus, the $x^{'} and y^{'}$ - coordinates become $x^{'} = (0) (\frac{\sqrt{3}}{2}) + (2) (\frac{1}{2})$
$= 0 + (1)$
=1
$y^{'} = - (0) (\frac{1}{2}) + (2) (\frac{\sqrt{3}}{2})$
$= 0 + (\sqrt{3})$
$= \sqrt{3}$
Therefore, the coordinates of the point the $x^{'} y^{'}$ - coordinate system are 1 and $\sqrt{3}$ , respectively.

The coordinates of the point in the x^{prime} y^{prime} - coordinate system with the given angle of rotation and the xy-coordinates.

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