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Find the length of the line segments whose endpoints have polar coordinate (7, 40 degres) and (15, 100 degrees).

 Dec 12, 2019

Best Answer 

 #1
avatar+24350 
+3

Find the length of the line segments whose endpoints have polar coordinate (7, 40 degres) and (15, 100 degrees).

 

\(\text{Let $P_1 =\dbinom{x_1}{y_1}=\dbinom{7\cos(40^\circ)}{7\sin(40^\circ) } $} \\ \text{Let $P_2 =\dbinom{x_2}{y_2}=\dbinom{15\cos(100^\circ)}{15\sin(100^\circ)} $} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{ \overline{P_1P_2} } &=& \mathbf{ \sqrt{ \left( x_2-x_1\right)^2 + \left(y_2-y_1 \right)^2 } } \\\\ &=& \sqrt{ \Big( 15\cos(100^\circ) - 7\cos(40^\circ)\Big)^2 + \Big( 15\sin(100^\circ) - 7\sin(40^\circ)\Big)^2 } \\ &=& \tiny{\sqrt{ 15^2\cos^2(100^\circ) -2*15*7\cos(100^\circ)\cos(40^\circ) + 7^2\cos^2(40^\circ) + 15^2\sin^2(100^\circ) -2*15*7\sin(100^\circ)\sin(40^\circ) + 7^2\sin^2(40^\circ) }} \\ &=& \tiny{\sqrt{ 7^2\Big(\underbrace{\sin^2(40^\circ)+\cos^2(40^\circ)}_{\small{=1}}\Big) + 15^2\Big(\underbrace{\sin^2(100^\circ)+\cos^2(100^\circ)}_{\small{=1}}\Big) -2*15*7\Big(\underbrace{\cos(100^\circ)\cos(40^\circ) + \sin(100^\circ)\sin(40^\circ)}_{\small{=\cos(100^\circ-40^\circ)}} \Big) }} \\ &=& \sqrt{ 7^2+ 15^2 -2*15*7\cos(100^\circ-40^\circ) } \\ &=& \sqrt{ 7^2+ 15^2 -2*15*7\cos(60^\circ) } \quad | \quad \cos(60^\circ)= \dfrac{1}{2} \\ &=& \sqrt{ 7^2+ 15^2 - 15*7 } \\ &=& \sqrt{ 169 } \\ &=& \mathbf{13} \\ \hline \end{array}\)

 

The length of the line is 13

 

laugh

 Dec 12, 2019
 #1
avatar+24350 
+3
Best Answer

Find the length of the line segments whose endpoints have polar coordinate (7, 40 degres) and (15, 100 degrees).

 

\(\text{Let $P_1 =\dbinom{x_1}{y_1}=\dbinom{7\cos(40^\circ)}{7\sin(40^\circ) } $} \\ \text{Let $P_2 =\dbinom{x_2}{y_2}=\dbinom{15\cos(100^\circ)}{15\sin(100^\circ)} $} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{ \overline{P_1P_2} } &=& \mathbf{ \sqrt{ \left( x_2-x_1\right)^2 + \left(y_2-y_1 \right)^2 } } \\\\ &=& \sqrt{ \Big( 15\cos(100^\circ) - 7\cos(40^\circ)\Big)^2 + \Big( 15\sin(100^\circ) - 7\sin(40^\circ)\Big)^2 } \\ &=& \tiny{\sqrt{ 15^2\cos^2(100^\circ) -2*15*7\cos(100^\circ)\cos(40^\circ) + 7^2\cos^2(40^\circ) + 15^2\sin^2(100^\circ) -2*15*7\sin(100^\circ)\sin(40^\circ) + 7^2\sin^2(40^\circ) }} \\ &=& \tiny{\sqrt{ 7^2\Big(\underbrace{\sin^2(40^\circ)+\cos^2(40^\circ)}_{\small{=1}}\Big) + 15^2\Big(\underbrace{\sin^2(100^\circ)+\cos^2(100^\circ)}_{\small{=1}}\Big) -2*15*7\Big(\underbrace{\cos(100^\circ)\cos(40^\circ) + \sin(100^\circ)\sin(40^\circ)}_{\small{=\cos(100^\circ-40^\circ)}} \Big) }} \\ &=& \sqrt{ 7^2+ 15^2 -2*15*7\cos(100^\circ-40^\circ) } \\ &=& \sqrt{ 7^2+ 15^2 -2*15*7\cos(60^\circ) } \quad | \quad \cos(60^\circ)= \dfrac{1}{2} \\ &=& \sqrt{ 7^2+ 15^2 - 15*7 } \\ &=& \sqrt{ 169 } \\ &=& \mathbf{13} \\ \hline \end{array}\)

 

The length of the line is 13

 

laugh

heureka Dec 12, 2019
 #2
avatar+109064 
+1

Nice, heureka....!!!!

 

cool cool cool

CPhill  Dec 12, 2019
 #3
avatar+24350 
+1

Thank you, CPhill !

 

laugh

heureka  Dec 13, 2019

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