What is the distance between the following polar coordinates?: $(7, \frac{5 π}{4}), (2, \frac{9 π}{8})$ $(7,(5pi)/4), (2,(9pi)/8)$

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What is the distance between the following polar coordinates?: $(7, \frac{5 π}{4}), (2, \frac{9 π}{8})$ $(7,(5pi)/4), (2,(9pi)/8)$

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Answer 1 · 2016-04-04T21:17:04

$P_{1} P_{2} = \sqrt{53 - 28 cos (\frac{π}{8})} \approx 5.209$ $P_1P_2 = sqrt(53-28cos((pi)/8)) ~~5.209$

Explanation:

$P_{1} P_{2} = \sqrt{r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} cos (θ_{2} - θ_{1})}$ $P_1P_2 = sqrt(r_1^2+r_2^2-2r_1r_2cos(theta_2-theta_1))$
$r_{1} = 7, θ_{1} = \frac{5 π}{4}; r_{2} = 2, θ_{2} = \frac{9 π}{8}$ $r_1 = 7, theta_1 =(5pi)/4;r_2 =2, theta_2 =(9pi)/8$
$P_{1} P_{2} = \sqrt{7^{2} + 2^{2} - 2 \cdot 7 \cdot 2 cos (\frac{9 π}{8} - \frac{5 π}{4})}$ $P_1P_2 = sqrt(7^2+2^2-2*7*2cos((9pi)/8-(5pi)/4))$
$P_{1} P_{2} = \sqrt{49 + 4 - 28 cos (- \frac{π}{8})}$ $P_1P_2 = sqrt(49+4-28cos(-(pi)/8)$
$P_{1} P_{2} = \sqrt{53 - 28 cos (\frac{π}{8})} \approx 5.209$ $P_1P_2 = sqrt(53-28cos((pi)/8)) ~~5.209$

Answer 2 · 2016-04-04T21:23:44

$s ≅ 5, 27$ $s~=5,27$

Explanation:

$r_{1} = 7$ $r_1=7$
$r_{2} = 2$ $r_2=2$
$θ_{1} = \frac{5 π}{4}$ $theta_1=(5pi)/4$
$θ_{2} = \frac{9 π}{8}$ $theta_2=(9pi)/8$
$θ_{2} - θ_{1} = \frac{9 π}{8} - \frac{5 π}{4} = \frac{9 π - 10 π}{8} = - \frac{π}{8}$ $theta_2-theta_1=(9pi)/8-(5pi)/4=(9pi-10pi)/8=-pi/8$

$cos (- \frac{π}{8}) = 0, 9$ $cos(-pi/8)=0,9$

$s = \sqrt{r_{1}^{2} + r_{2}^{2} - 2 \cdot r_{1} \cdot r_{2} \cdot cos (θ_{2} - θ_{1})}$ $s=sqrt(r_1^2+r_2^2-2*r_1*r_2*cos(theta_2-theta_1))$

$s = \sqrt{7^{2} + 2^{2} - 2 \cdot 7 \cdot 2 \cdot 0, 9}$ $s=sqrt(7^2+2^2-2*7*2*0,9)$

$s = \sqrt{49 + 4 - 28 \cdot 0, 9}$ $s=sqrt(49+4-28*0,9)$

$s = \sqrt{53 - 25.2}$ $s=sqrt(53-25.2)$

$s = \sqrt{27, 8}$ $s=sqrt(27,8)$
$s ≅ 5.27$ $s~=5.27$

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What is the distance between the following polar coordinates?: $(7, \frac{5 π}{4}), (2, \frac{9 π}{8})$ $(7,(5pi)/4), (2,(9pi)/8)$

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What is the distance between the following polar coordinates?: $(7, \frac{5 π}{4}), (2, \frac{9 π}{8})$ $(7,(5pi)/4), (2,(9pi)/8)$