Find the rectangular coordinate of the point $\left( {5,300} \right).$
Answer
627k+ views
Hint:To convert Polar coordinates$\left( {r,\theta } \right)$to rectangular coordinates $\left( {x,y} \right)$,
we have the equation:
$
x = r\cos \theta \\
y = r\sin \theta \\
$
So by using the above equation and substituting the needed values we can find the rectangular coordinate corresponding to polar coordinate $\left( {5,300} \right).$
Complete step by step solution:
Given
$\left( {5,300} \right).....................\left( i \right)$
We know that rectangle coordinates are the Cartesian coordinates seen in the Cartesian plane which is represented by $\left( {x,y} \right)$and polar coordinates give the position of a point in a plane by using the length$r$and the angle made to the fixed point $\theta $, and is represented by $\left( {r,\theta }
\right).$
We know that (i) which is a polar coordinate is to be converted to a rectangular coordinate.
For that we can use the formula:
$
x = r\cos \theta .................\left( {ii} \right) \\
y = r\sin \theta ..................\left( {iii} \right) \\
$
So by substituting the values of $r\,\,{\text{and}}\,\,\theta $ in the equation (ii) and (iii) we can find the
values of $x\,\,{\text{and}}\,\,y.$
Now we know that on comparing (i) we can write:
$r = 5$
$\theta = 300 = \left( {2\pi - \dfrac{\pi }{3}} \right)$
i.e. changing $\theta $ from degrees to radians.
Now substituting the values of $r\,\,{\text{and}}\,\,\theta $ in the equation (ii) and (iii), we get:
$
\Rightarrow x = r\cos \theta = 5 \times \cos \left( {2\pi - \dfrac{\pi }{3}} \right) \\
\,\,\,\,\,\,\,\,\,x = 5 \times \dfrac{1}{2} \\
$
$ \Rightarrow x = \dfrac{5}{2}.......................\left( {iv} \right)$
Now for finding y:
$
\Rightarrow y = r\sin \theta \\
\,\,\,\,\,\,\,\,\,\,\, = 5 \times \sin \left( {2\pi - \dfrac{\pi }{3}} \right) \\
\,\,\,\,\,\,\,\,\,\,\, = 5 \times - \dfrac{{\sqrt 3 }}{2} \\
\Rightarrow y = - \dfrac{{5\sqrt 3 }}{2}..................\left( v \right) \\
$
So from (iv) and (v) we have got the values of $x = \dfrac{5}{2}\,\,{\text{and}}\,\,y = - \dfrac{{5\sqrt 3
}}{2}$, which are our rectangular coordinates.
Therefore the rectangular coordinate of the point $\left( {5,300} \right)$ is $\left( {\dfrac{5}{2}, -
\dfrac{{5\sqrt 3 }}{2}} \right)$.
Note: We know that to convert a polar coordinate$\left( {r,\theta } \right)$to a rectangular coordinate$\left(
{x,y} \right)$, we can use the formula:
$
x = r\cos \theta \\
y = r\sin \theta \\
$
In a similar manner convert rectangular coordinate$\left( {x,y} \right)$to a polar coordinate $\left( {r,\theta } \right)$, we can use the formula:
$
r = \sqrt {\left( {{x^2} + {y^2}} \right)} \\
\theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) \\
$
Also while choosing $\theta $ it’s better to choose it in radians since when $\theta $ is in radians the calculations become much easier.
we have the equation:
$
x = r\cos \theta \\
y = r\sin \theta \\
$
So by using the above equation and substituting the needed values we can find the rectangular coordinate corresponding to polar coordinate $\left( {5,300} \right).$
Complete step by step solution:
Given
$\left( {5,300} \right).....................\left( i \right)$
We know that rectangle coordinates are the Cartesian coordinates seen in the Cartesian plane which is represented by $\left( {x,y} \right)$and polar coordinates give the position of a point in a plane by using the length$r$and the angle made to the fixed point $\theta $, and is represented by $\left( {r,\theta }
\right).$
We know that (i) which is a polar coordinate is to be converted to a rectangular coordinate.
For that we can use the formula:
$
x = r\cos \theta .................\left( {ii} \right) \\
y = r\sin \theta ..................\left( {iii} \right) \\
$
So by substituting the values of $r\,\,{\text{and}}\,\,\theta $ in the equation (ii) and (iii) we can find the
values of $x\,\,{\text{and}}\,\,y.$
Now we know that on comparing (i) we can write:
$r = 5$
$\theta = 300 = \left( {2\pi - \dfrac{\pi }{3}} \right)$
i.e. changing $\theta $ from degrees to radians.
Now substituting the values of $r\,\,{\text{and}}\,\,\theta $ in the equation (ii) and (iii), we get:
$
\Rightarrow x = r\cos \theta = 5 \times \cos \left( {2\pi - \dfrac{\pi }{3}} \right) \\
\,\,\,\,\,\,\,\,\,x = 5 \times \dfrac{1}{2} \\
$
$ \Rightarrow x = \dfrac{5}{2}.......................\left( {iv} \right)$
Now for finding y:
$
\Rightarrow y = r\sin \theta \\
\,\,\,\,\,\,\,\,\,\,\, = 5 \times \sin \left( {2\pi - \dfrac{\pi }{3}} \right) \\
\,\,\,\,\,\,\,\,\,\,\, = 5 \times - \dfrac{{\sqrt 3 }}{2} \\
\Rightarrow y = - \dfrac{{5\sqrt 3 }}{2}..................\left( v \right) \\
$
So from (iv) and (v) we have got the values of $x = \dfrac{5}{2}\,\,{\text{and}}\,\,y = - \dfrac{{5\sqrt 3
}}{2}$, which are our rectangular coordinates.
Therefore the rectangular coordinate of the point $\left( {5,300} \right)$ is $\left( {\dfrac{5}{2}, -
\dfrac{{5\sqrt 3 }}{2}} \right)$.
Note: We know that to convert a polar coordinate$\left( {r,\theta } \right)$to a rectangular coordinate$\left(
{x,y} \right)$, we can use the formula:
$
x = r\cos \theta \\
y = r\sin \theta \\
$
In a similar manner convert rectangular coordinate$\left( {x,y} \right)$to a polar coordinate $\left( {r,\theta } \right)$, we can use the formula:
$
r = \sqrt {\left( {{x^2} + {y^2}} \right)} \\
\theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) \\
$
Also while choosing $\theta $ it’s better to choose it in radians since when $\theta $ is in radians the calculations become much easier.
Recently Updated Pages
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

