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

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

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Answer 1 · 2015-03-03T18:56:04

The distance is $2.88$ $2.88$ .

As you can see from this drawing:

Geogebra

$A and B$ $A and B$ in polar coordinates are $A (O A, α)$ $A(OA,alpha)$ and $B (O B, β)$ $B(OB,beta)$ .

I have named $γ$ $gamma$ the angle between the two vectors $v_{1}$ $v_1$ and $v_{2}$ $v_2$ , and, as you can easily see $γ = α - β$ $gamma=alpha-beta$ .
(It's not important if we do $α - β$ $alpha-beta$ or $β - α$ $beta-alpha$ because, at the end, we will calculate the cosine of $γ$ $gamma$ and $cos γ = cos (- γ)$ $cosgamma=cos(-gamma)$ ).

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

So we can use the cosine theorem, that says:

$a^{2} = b^{2} + c^{2} - 2 b c cos α$ $a^2=b^2+c^2-2bc cosalpha$ ,

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

So, in our case:

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

$=sqrt(9+1/4-3cos(71))~=2,88$ .

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

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