# What is the distance between (-6 , pi/2 ) and (5, pi/6 )?

Apr 10, 2016

$\setminus \sqrt{91}$ units.

#### Explanation:

Since the coordinates are polar coordinates, we imagine a triangle $O A B$ with $O$ at the origin, $A$ at $\left(- 6 , \setminus \frac{\pi}{2}\right)$ and $B$ at $\left(5 , \setminus \frac{\pi}{6}\right)$. We will use the Law of Cosines to get the length of $A B$.

First convert side $O A$ to a positive length by writing its coordinates as $\left(+ 6 , - \setminus \frac{\pi}{2}\right)$, changing the sign of the radius and compensating by subtracting $\setminus \pi$ from the angle.

So $O A = 6$ and $O B = 5$. Next we need angle $O$ which is the difference between the angular coordinates after we have made the radial coordinates positive (see above). Thus

$\setminus \frac{\pi}{2} - \left(- \setminus \frac{\pi}{6}\right) = 2 \setminus \frac{\pi}{3}$.

Now apply the Law of Cosines:

${\left(A B\right)}^{2} = {\left(O A\right)}^{2} + {\left(O B\right)}^{2} - 2 \left(O A\right) \left(O B\right) \setminus \cos \left(\angle O\right)$

$= {6}^{2} + {5}^{2} - 2 \left(6\right) \left(5\right) \setminus \cos \left(2 \setminus \frac{\pi}{3}\right)$

As $\setminus \cos \left(2 \setminus \frac{\pi}{3}\right) = - \frac{1}{2}$ this gives ${\left(A B\right)}^{2} = 91$.