document.write( "Question 253576: Hi :o)
\n" ); document.write( "I hope you guys can help me with this. I think I now understand how to work out a resultant vector using the coordinate system, but this question has stumped me a little. I can't get a picture of it - it seems to be 3D, is that right?
\n" ); document.write( "Please help...\r
\n" ); document.write( "\n" ); document.write( "A mine shaft goes due west 75 m from the opening at an angle of 25° below
\n" ); document.write( "the horizontal surface. It then becomes horizontal and turns 30° north of west
\n" ); document.write( "and continues for another 45 m. What is the displacement of the end of the
\n" ); document.write( "tunnel from the opening.\r
\n" ); document.write( "\n" ); document.write( "Thanks SOOOOO much :o) (Happy New Year)
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Algebra.Com's Answer #185916 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
A mine shaft goes due west 75 m from the opening at an angle of 25° below
\n" ); document.write( "the horizontal surface. It then becomes horizontal and turns 30° north of west
\n" ); document.write( "and continues for another 45 m. What is the displacement of the end of the
\n" ); document.write( "tunnel from the opening.
\n" ); document.write( "-------------
\n" ); document.write( "If you know how to graph in 3D,
\n" ); document.write( "the start point is (0,0,0)
\n" ); document.write( "the turning point is (-75cos(25), 0 ,-75sin(25))
\n" ); document.write( "the final position is (-75cos(25) ,-45,-75sin(25))
\n" ); document.write( "-------------------------------
\n" ); document.write( "Distance from start to finish:
\n" ); document.write( "d = sqrt((75cos(25))^2 + (45)^2 + (75sin(25))^2)
\n" ); document.write( "d = sqrt(5526.47+2025+98.53)
\n" ); document.write( "d = 87.46 meters
\n" ); document.write( "==========================
\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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