Lesson Objectives
  • Learn how to find the angle between two polar points
  • Learn how to find the distance between two polar points

How to Find the Distance Between Two Polar Points


In this lesson, we will learn how to find the distance between points on the polar plane. Previously, we have learned how to find the distance between two points (x1, y1), (x2, y2) on the rectangular plane using the distance formula: $$d=\sqrt{(x_{2}- x_{1})^{2}+ (y_{2}- y_{1})^2}$$ Let's suppose we were asked to find the distance between two points given with polar coordinates: $$(r_{1}, θ_{1}), (r_{2}, θ_{2})$$ We could convert these polar coordinates to rectangular coordinates and then use the distance formula, however, there is a simpler approach. Let's consider finding the distance between the two polar points: $$(4, 105°), (3, 225°)$$ Let's plot our two points on the polar grid: Graphing the points (4, 105°), (3, 225°) Let's first consider the angle between the two points: Finding the angle between the points (4, 105°), (3, 225°) We need to rotate 105° counterclockwise to get from the polar axis to the first point (4, 105°). Additionally, we need to rotate 225° counterclockwise to get from the polar axis to the second point. The angle between the two points (120°) would be found by subtracting the larger angle 225° minus the smaller angle 105°: Finding the angle between the points (4, 105°), (3, 225°) Now that we have found the angle between the two points, let's draw a line segment connecting these two points. The length of this line segment is the distance between the two points. Graphing the points (4, 105°), (3, 225°) with a line segment connecting the two Now, let's form a triangle by drawing a line segment from the pole to (4, 105°) and another from the pole to (3, 225°): Creating an oblique triangle Consider that we have SAS, which allows us to use the law of cosines to find our missing side length.
In any triangle ABC, with sides a, b, and c: $$a^{2}=b^{2}+ c^{2}- 2bc\hspace{.1em}\text{cos}\hspace{.1em}A$$ Let's call our unknown side a, this will be the distance between the two polar points. The other two known sides will be b and c: $$a^{2}=4^{2}+ 3^{2}- 2(4)(3) \cdot \text{cos}120°$$ $$a^{2}=16 + 9 - \left(24 \cdot -\frac{1}{2}\right)$$ $$a^{2}=25 - (-12)$$ $$a^{2}=37$$ $$a=\sqrt{37}$$ This tells us the distance between the two points (4, 105°) and (3, 225°) is $\sqrt{37}$.

Formula for Finding the Distance Between Two Polar Points

Let's now generalize our results into a nice formula. $$(r_{1}, θ_{1}), (r_{2}, θ_{2})$$ We will just plug into the law of cosines with r1 and r2 representing our two known sides and the angle between the two sides will be given as (θ2 - θ1): $$a^{2}=b^{2}+ c^{2}- 2bc\hspace{.1em}\text{cos}\hspace{.1em}A$$ We will use d for distance instead of a, the unknown side length: $$d^{2}=(r_{1})^{2}+ (r_{2})^{2}- 2(r_{1})(r_{2}) \cdot \text{cos}(θ_{2}- θ_{1})$$ $$d=\sqrt{(r_{1})^{2}+ (r_{2})^{2}- 2(r_{1})(r_{2}) \cdot \text{cos}(θ_{2}- θ_{1})}$$ Let's look at a few examples.
Example #1: Find the distance between the polar points. $$(2, 120°), (4, 210°)$$ Let's just use the left point as the first point and the right point as the second point. Note, the labeling doesn't matter. We will get the same answer either way: $$d=\sqrt{2^{2}+ 4^{2}- 2(2)(4) \cdot \text{cos}(210° - 120°)}$$ $$d=\sqrt{4 + 16 - (16 \cdot \text{cos}(90°))}$$ $$d=\sqrt{20 - (16 \cdot 0)}$$ $$d=\sqrt{20}$$ $$d=2\sqrt{5}$$ If we reversed the labeling, we would end up with cos(-90°).
Recall the negative angle identity for cosine: $$\text{cos}(-θ)=\text{cos}\hspace{.1em}θ$$ Therefore, we would have the same answer either way.
Example #2: Find the distance between the polar points. $$\left(3, \frac{7π}{6}\right), \left(2, \frac{π}{2}\right)$$ Let's again use the left point as the first point and the right point as the second point: $$d=\sqrt{3^{2}+ 2^{2}- 2(3)(2) \cdot \text{cos}\left(\frac{π}{2}- \frac{7π}{6}\right)}$$ $$d=\sqrt{9 + 4 - \left(12 \cdot \text{cos}\left(\frac{π}{2}- \frac{7π}{6}\right)\right)}$$ $$d=\sqrt{13 - \left(12 \cdot \text{cos}\left(\frac{3π}{6}- \frac{7π}{6}\right)\right)}$$ $$d=\sqrt{13 - \left(12 \cdot \text{cos}\left(-\frac{4π}{6}\right)\right)}$$ $$d=\sqrt{13 - \left(12 \cdot \text{cos}\left(-\frac{2π}{3}\right)\right)}$$ Use the negative angle identity for cosine: $$d=\sqrt{13 - \left(12 \cdot \text{cos}\left(\frac{2π}{3}\right)\right)}$$ $$d=\sqrt{13 - \left(12 \cdot -\frac{1}{2}\right)}$$ $$d=\sqrt{13 - (-6)}$$ $$d=\sqrt{19}$$ When we are given a problem with a negative r-value, a negative angle, or an angle that is larger than 360°, we can use the formula as given. If it helps, you can also write the point in a different form with a positive radius and positive angle between 0° (inclusive) and 360° (exclusive). Let's look at a few examples.
Example #3: Find the distance between the polar points. $$(4, -30°), (4, 210°)$$ For the first point, we have a negative angle. We can work with the angle as is or we can find a coterminal angle by adding 360°. This will locate another form of the same point. Either way, the answer will be the same. Let's try it both ways and think about why this works.
Plug into the formula, use the left point as the first point and the right point as the second point: $$d=\sqrt{4^{2}+ 4^{2}- 2(4)(4) \cdot \text{cos}(210 - (-30))}$$ $$d=\sqrt{32 - 32 \cdot \text{cos}(240)}$$ $$d=\sqrt{32 - \left(32 \cdot -\frac{1}{2}\right)}$$ $$d=\sqrt{32 - (-16)}$$ $$d=\sqrt{48}$$ $$d=4\sqrt{3}$$ Let's now work the problem in a different way. Let's find a coterminal angle for -30° by adding 360°: $$-30° + 360°=330°$$ Our new polar coordinates locate the same point but now we have an angle between 0° (inclusive) and 360° (exclusive). $$(4, 330°), (4, 210°)$$ Plug into the formula, use the left point as the first point and the right point as the second point: $$d=\sqrt{4^{2}+ 4^{2}- 2(4)(4) \cdot \text{cos}(210° - 330°)}$$ $$d=\sqrt{16 + 16 - 32 \cdot \text{cos}(-120°)}$$ Use the negative angle identity for cosine: $$d=\sqrt{32 - (32 \cdot \text{cos}(120°))}$$ $$d=\sqrt{32 - \left(32 \cdot -\frac{1}{2}\right)}$$ $$d=\sqrt{32 - (-16)}$$ $$d=\sqrt{48}$$ $$d=4\sqrt{3}$$ You can see that we get the same answer either way and might be wondering why? Let's take a look at our points on the polar grid: plotting (4, -30°), (4, 210°) Let's think about a simple identity: $$\text{cos}(360° - θ)=\text{cos}\hspace{.1em}θ$$ We can prove this using the difference identity for cosine: $$\text{cos}(360° - θ)=\text{cos}\hspace{.1em}360° \hspace{.1em}\text{cos}\hspace{.1em}θ + \text{sin}\hspace{.1em}360° \hspace{.1em}\text{sin}\hspace{.1em}θ$$ $$\text{cos}(360° - θ)=1 \hspace{.1em}\text{cos}\hspace{.1em}θ + 0$$ $$\text{cos}(360° - θ)=\hspace{.1em}\text{cos}\hspace{.1em}θ$$ This tells us that it doesn't matter if we find the angle between the polar points or the angle on the outside. The cosine value will be the same either way, so we can use the formula with any negative angles that we come across.
Example #4: Find the distance between the polar points. $$(-3, -165°), (4, 45°)$$ For the first point, we have a negative r-value and a negative angle. We can work with this point as given or change it over by adding 180° to the angle and changing the r-value to positive. $$(3, 15°), (4, 45°)$$ Plug into the formula, use the left point as the first point and the right point as the second point: $$d=\sqrt{3^{2}+ 4^{2}- 2(3)(4) \cdot \text{cos}(45° - 15°)}$$ $$d=\sqrt{9 + 16 - 24 \cdot \text{cos}(30°)}$$ $$d=\sqrt{25 - \left(24 \cdot \frac{\sqrt{3}}{2}\right)}$$ $$d=\sqrt{25 - 12\sqrt{3}}$$ If we wanted to skip the conversion, our formula would work just fine. Let's plug in the polar points as given.
Plug into the formula, use the left point as the first point and the right point as the second point: $$d=\sqrt{(-3)^{2}+ 4^{2}- 2(-3)(4) \cdot \text{cos}(45 - (-165°))}$$ $$d=\sqrt{9 + 16 + 24 \cdot \text{cos}(210°)}$$ $$d=\sqrt{25 + \left(24 \cdot -\frac{\sqrt{3}}{2}\right)}$$ $$d=\sqrt{25 - 12 \sqrt{3}}$$ Again, let's pause for a minute and think about why this works. When we have a negative r-value, this changes our sign in the middle part to positive. This is countered by the fact that changing to a positive r-value means the angle is increased by 180°. Let's think about the following identity. $$\text{cos}(θ + 180°)=-\text{cos}\hspace{.1em}θ$$ We can prove this using the sum identity for cosine: $$\text{cos}(θ + 180°)=-\text{cos}\hspace{.1em}θ$$ $$\text{cos}(θ + 180°)=\text{cos}\hspace{.1em}θ \hspace{.1em}\text{cos}\hspace{.1em}180° - \text{sin}\hspace{.1em}θ \hspace{.1em}\text{sin}\hspace{.1em}180°$$ $$\text{cos}(θ + 180°)=-1 \cdot \text{cos}\hspace{.1em}θ - 0$$ $$\text{cos}(θ + 180°)=-\text{cos}\hspace{.1em}θ$$

Skills Check:

Example #1

Find the distance between the polar points. $$\left(1, \frac{4π}{3}\right), \left(1, \frac{13π}{12}\right)$$

Please choose the best answer.

A
$$2\sqrt{5}$$
B
$$2$$
C
$$\sqrt{2-\sqrt{2}}$$
D
$$\sqrt{7}$$
E
$$\sqrt{5 - \sqrt{2}}$$

Example #2

Find the distance between the polar points. $$(-2, 15°), (-4, 240°)$$

Please choose the best answer.

A
$$2\sqrt{13}$$
B
$$2\sqrt{5}$$
C
$$2\sqrt{5 + 2\sqrt{2}}$$
D
$$\sqrt{2 - \sqrt{2}}$$
E
$$\sqrt{25 + 12\sqrt{2}}$$

Example #3

Find the distance between the polar points. $$(3, -90°), (-1, 135°)$$

Please choose the best answer.

A
$$\sqrt{10 - 3\sqrt{2}}$$
B
$$\sqrt{5 + 2\sqrt{2}}$$
C
$$\sqrt{2}$$
D
$$\sqrt{2 + \sqrt{2}}$$
E
$$\sqrt{3 + \sqrt{19}}$$
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