# How do I find the distance between the points with polar coordinates (3, 120^circ) and (0.5, 49^circ)?

Mar 3, 2015

The distance is $2.88$.

As you can see from this drawing: $A \mathmr{and} B$ in polar coordinates are $A \left(O A , \alpha\right)$ and $B \left(O B , \beta\right)$.

I have named $\gamma$ the angle between the two vectors ${v}_{1}$ and ${v}_{2}$, and, as you can easily see $\gamma = \alpha - \beta$.
(It's not important if we do $\alpha - \beta$ or $\beta - \alpha$ because, at the end, we will calculate the cosine of $\gamma$ and $\cos \gamma = \cos \left(- \gamma\right)$).

We know, of the triangle $A O B$, two sides and the angle between them and we have to find the segment $A B$, that is the distance between $A$ and $B$.

So we can use the cosine theorem, that says:

${a}^{2} = {b}^{2} + {c}^{2} - 2 b c \cos \alpha$,

where $a , b , c$ are the three sides of a triangle and $\alpha$ is the angle between $b$ and $c$.

So, in our case:

AB=sqrt(3^2+(1/2)^2-2*3*1/2*cos(120°-49°))=

$= \sqrt{9 + \frac{1}{4} - 3 \cos \left(71\right)} \cong 2 , 88$.