What is the distance between $(- 6, \frac{π}{2})$ $(-6 , pi/2 )$ and $(1, \frac{π}{6})$ $(1 , pi/6 )$ ?

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Answer 1 · 2018-01-27T11:59:58

$\sqrt{43}$ $sqrt(43)$

This problem is more easily done in rectangular coordinates:

$(- 6, \frac{π}{2}) \to (0, - 6)$ $(-6,pi/2) to (0,-6)$
$(1, \frac{π}{6}) \to (\frac{\sqrt{3}}{2}, \frac{1}{2})$ $(1,pi/6) to (sqrt(3)/2, 1/2)$

Now we can just use the distance formula:

$\sqrt{{(\frac{\sqrt{3}}{2} - 0)}^{2} + {(\frac{1}{2} - (- 6))}^{2}}$ $sqrt((sqrt(3)/2 - 0)^2+(1/2-(-6))^2)$

$= \sqrt{\frac{3}{4} + {(\frac{13}{2})}^{2}}$ $=sqrt(3/4+(13/2)^2)$

$= \sqrt{\frac{3}{4} + \frac{169}{4}}$ $=sqrt(3/4+169/4)$

$= \sqrt{\frac{172}{4}}$ $=sqrt(172/4)$

$\sqrt{43}$ $sqrt(43)$

More generally, if we rewrite the first point as $(6, - \frac{π}{2})$ $(6, -pi/2)$ , then we can use the Law of Cosines, [as shown in this

Answer 2 · 2018-01-27T12:37:18

Distance between points is $6.56$ $6.56$ unit .

Polar coordinates of point A is $r_{1} = - 6, θ_{1} = \frac{π}{2}$ $r_1= -6 ,theta_1=pi/2$

Polar coordinates of point B is $r_{2} = 1, θ_{2} = \frac{π}{6}$ $r_2=1, theta_2=pi/6$

Distance between points $A and B$ $A and B$ is

$D = \sqrt{r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} cos (θ 1 - θ 2)}$ $D= sqrt(r_1^2+r_2^2-2r_1r_2cos(theta1-theta2)$ or

$D = \sqrt{36 + 1 + 12 \cdot cos (\frac{π}{2} - \frac{π}{6})} \approx 6.56 (2 d p)$ $D = sqrt(36+1+12*cos (pi/2-pi/6)) ~~ 6.56(2dp)$ unit

Distance between points is $6.56$ $6.56$ unit [Ans]

What is the distance between (−6,π2)(-6 , pi/2 ) and (1,π6)(1 , pi/6 )?