Question

How do you find the distance between `P_1(2.5,pi/6)` and `P_2(-3,-pi/4)` on the polar plane?

Answer 1

Please see the explanation for steps leading to the answer: `c ~~ 4.374`

Explanation:

Change the radius of `P_2` from negative to positive by adding `pi` to the angle:

`P_2(3, (3pi)/4)`

The origin and the two points form a triangle with side `a= 2.5`, side `b = 3` and the angle between them `theta = (3pi)/4 - pi/6 = (14pi)/24 = (7pi)/12`. Therefore, we can use the Law of Cosines to find the distance, c, between them:

`c = sqrt(a^2 + b^2 - 2(a)(b)cos(theta))`

`c = sqrt(2.5^2 + 3^2 - 2(2.5)(3)cos((7pi)/12))`

`c = sqrt(2.5^2 + 3^2 - 2(2.5)(3)cos((7pi)/12))`

`c ~~ 4.374`