How do you find the rectangular coordinates given $ (6,150^\circ ) $ ?
Answer
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Hint: In this question, we are given an ordered pair, it is also known as the coordinates of a point. We see that the first one is an integer and the second one is an angle, so the given coordinates are of the form $ (r,\theta ) $ . There are two types of coordinates for plotting a point on the graph paper namely rectangular coordinate system and polar coordinate system. The given coordinates are the polar coordinates and we have to find the rectangular coordinates.
A right-angled triangle is formed by x, y and r, where x is the base, y is the height and r is the hypotenuse of the right-angled triangle, so by Pythagoras theorem, we get – $ {x^2} + {y^2} = {r^2} $ and by trigonometry we get –
$
\cos \theta = \dfrac{{base}}{{hypotenuse}} = \dfrac{x}{{\sqrt {{x^2} + {y^2}} }} = \dfrac{x}{r} \\
\Rightarrow x = r\cos \theta \;
$
And similarly $ y = r\sin \theta $
Using this information, we can find the value of x and y.
Complete step by step solution:
We are given the polar coordinates are $ (6,150^\circ ) $ so we get $ r = 6 $ and $ \theta = 150^\circ $
We know that
$
x = r\cos \theta \\
\Rightarrow x = 6\cos (150^\circ ) \\
\Rightarrow x = 6\cos (180^\circ - 30^\circ ) \;
$
We know $ \cos (180 - x) = - \cos x $
$
\Rightarrow x = 6( - \cos 30^\circ ) \\
\Rightarrow x = - 6 \times \dfrac{{\sqrt 3 }}{2} \\
\Rightarrow x = - 3\sqrt 3 \;
$
$
y = r\sin \theta \\
\Rightarrow y = 6\sin (150^\circ ) \\
\Rightarrow y = 6(\sin 180^\circ - 30^\circ ) \;
$
We know that $ \sin (180 - x) = \sin x $
$
\Rightarrow y = 6\sin 30^\circ \\
\Rightarrow y = 6 \times \dfrac{1}{2} \\
\Rightarrow y = 3 \;
$
Hence when the polar coordinates are $ (6,150^\circ ) $ , the rectangular coordinates are $ ( - 3\sqrt 3 ,3) $
So, the correct answer is “ $ ( - 3\sqrt 3 ,3) $ ”.
Note: The polar coordinate system is of the form $ (r,\theta ) $ where r is the distance of the point from the origin and $ \theta $ is the counter-clockwise angle between the line joining the point and the origin and the x-axis. The rectangular coordinate system is of the form $ (x,y) $ and is the most commonly used coordinate system, where x is the distance of this point from the y-axis and y is the distance of the point from the x-axis.
A right-angled triangle is formed by x, y and r, where x is the base, y is the height and r is the hypotenuse of the right-angled triangle, so by Pythagoras theorem, we get – $ {x^2} + {y^2} = {r^2} $ and by trigonometry we get –
$
\cos \theta = \dfrac{{base}}{{hypotenuse}} = \dfrac{x}{{\sqrt {{x^2} + {y^2}} }} = \dfrac{x}{r} \\
\Rightarrow x = r\cos \theta \;
$
And similarly $ y = r\sin \theta $
Using this information, we can find the value of x and y.
Complete step by step solution:
We are given the polar coordinates are $ (6,150^\circ ) $ so we get $ r = 6 $ and $ \theta = 150^\circ $
We know that
$
x = r\cos \theta \\
\Rightarrow x = 6\cos (150^\circ ) \\
\Rightarrow x = 6\cos (180^\circ - 30^\circ ) \;
$
We know $ \cos (180 - x) = - \cos x $
$
\Rightarrow x = 6( - \cos 30^\circ ) \\
\Rightarrow x = - 6 \times \dfrac{{\sqrt 3 }}{2} \\
\Rightarrow x = - 3\sqrt 3 \;
$
$
y = r\sin \theta \\
\Rightarrow y = 6\sin (150^\circ ) \\
\Rightarrow y = 6(\sin 180^\circ - 30^\circ ) \;
$
We know that $ \sin (180 - x) = \sin x $
$
\Rightarrow y = 6\sin 30^\circ \\
\Rightarrow y = 6 \times \dfrac{1}{2} \\
\Rightarrow y = 3 \;
$
Hence when the polar coordinates are $ (6,150^\circ ) $ , the rectangular coordinates are $ ( - 3\sqrt 3 ,3) $
So, the correct answer is “ $ ( - 3\sqrt 3 ,3) $ ”.
Note: The polar coordinate system is of the form $ (r,\theta ) $ where r is the distance of the point from the origin and $ \theta $ is the counter-clockwise angle between the line joining the point and the origin and the x-axis. The rectangular coordinate system is of the form $ (x,y) $ and is the most commonly used coordinate system, where x is the distance of this point from the y-axis and y is the distance of the point from the x-axis.
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