What is the distance between $(1, \frac{3 π}{4})$ $(1 ,(3 pi)/4 )$ and $(2, \frac{15 π}{8})$ $(2 , (15 pi )/8 )$ ?

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What is the distance between $(1, \frac{3 π}{4})$ $(1 ,(3 pi)/4 )$ and $(2, \frac{15 π}{8})$ $(2 , (15 pi )/8 )$ ?

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Answer 1 · 2016-12-02T17:40:35

$d \approx 2.95$ $d~~2.95$

Explanation:

Each point on a polar plane is represented by the ordered pair $(r, θ)$ $(r,theta)$ .

So lets call the coordinates of $P_{1}$ $P_1$ as $(r_{1}, θ_{1})$ $(r_1,theta_1)$ and coordinates of $P_{2}$ $P_2$ as $(r_{2}, θ_{2})$ $(r_2,theta_2)$ . To find the distance between two points on a polar plane use the formula $d = \sqrt{{(r_{1})}_{1}^{2} + {(r_{2})}_{2}^{2} - 2 r_{1} r_{2} cos (θ_{2} - θ_{1})}$ $d=sqrt((r_1) ^2+(r_2)^2-2r_1r_2cos(theta_2-theta_1))$

Thererfore using the points $(1, \frac{3 π}{4})$ $(1,(3pi)/4)$ and $(2, \frac{15 π}{8})$ $(2,(15pi)/8)$ , and the formula

$d = \sqrt{{(r_{1})}_{1}^{2} + {(r_{2})}_{2}^{2} - 2 r_{1} r_{2} cos (θ_{2} - θ_{1})}$ $d=sqrt((r_1) ^2+(r_2)^2-2r_1r_2cos(theta_2-theta_1))$

we have

$d = \sqrt{{(1)}^{2} + {(2)}^{2} - 2 \cdot 1 \cdot 2 cos (\frac{15 π}{8} - \frac{3 π}{4})}$ $d=sqrt((1)^2+(2)^2-2*1*2cos((15pi)/8-(3pi)/4))$

$:. d~~2.95$

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What is the distance between $(1, \frac{3 π}{4})$ $(1 ,(3 pi)/4 )$ and $(2, \frac{15 π}{8})$ $(2 , (15 pi )/8 )$ ?

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Explanation:

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