Question

What are the polar coordinates of point P = (3,8)?
1. $\left( {8.94,28.6^\circ } \right)$
2. $\left( {9.4,89.4^\circ } \right)$
3. $\left( {7.04,130.6^\circ } \right)$
4. $\left( {8.54,69.4^\circ } \right)$

Answer 1

Hint: In this question we have to find the polar coordinates from the given coordinates. So, we will convert the given coordinates in the form of $\left( {r,\theta } \right)$ where r is the magnitude and $\theta $ is the angle in polar form. Now, we will consider a position vector of the form $\vec r = x\hat i + y\hat j$ where angle can be taken out as $\theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$ and $r = \sqrt {{x^2} + {y^2}} $ . This will be used while solving the question.

Complete step-by-step solution:
Point P = (3,8) where according to the cartesian plane 3 is equal to x and 8 is equal to y.
Now, we will convert these coordinates into polar coordinates.
The polar coordinates are $\left( {r,\theta } \right)$. Where $r$ is the magnitude and $\theta $ is the angle in polar form.
Now, we will find the value of r vector,
$\vec r = x\hat i + y\hat j$ , here $\hat i$ and $\hat j$ are unit vectors along the x-axis and y-axis.
Substituting the values in the above equation.
$\vec r = 3\hat i + 8\hat j$
Taking magnitude on both sides.
$\left| {\vec r} \right| = \left| {3\hat i} \right| + \left| {8\hat j} \right|$
$r = \sqrt {{{\left( 3 \right)}^2} + {{\left( 8 \right)}^2}} $
$\Rightarrow r = \sqrt {9 + 64} $
$\Rightarrow r = \sqrt {73} $
$\Rightarrow r = 8.54$
Now, we will find the value of $\theta $ .
$\theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$
Substituting the values in the above equation.
$\theta = {\tan ^{ - 1}}\left( {\dfrac{8}{3}} \right)$
$\Rightarrow \theta = 69.4^\circ $
The polar coordinates of given coordinates is $\left( {8.54,69.4^\circ } \right)$ .
So, option (4) is the correct answer.

Note: To find the value of ${\tan ^{ - 1}}\left( {\dfrac{8}{3}} \right)$ , solve the trigonometric equation by isolating the function and taking the inverse. The trick point here was consideration of the fact that the given points are lying on the first quadrant as the x and y coordinates of the given points are both negative and thus the angle obtained has to be taken into the first quadrant only.