document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1172269>Question 1172269</A>: <I><SPAN STYLE=\"word-break: break-all;\"> An airplane is traveling at a speed of 200 miles/hour with a bearing of 240°. The wind velocity is 60 miles/hour at a bearing of 25°. What are the plane's actual speed and direction angle?\r<BR>\n" );
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document.write( "a. The speed is 140.33 miles/hour, and the direction angle is 237.87°.<BR>\n" );
document.write( "b. The speed is 140.73 miles/hour, and the direction angle is 197.15°.<BR>\n" );
document.write( "c. The speed is 154.73 miles/hour, and the direction angle is 77.85°.<BR>\n" );
document.write( "d. The speed is 154.73 miles/hour, and the direction angle is 197.15°.\r<BR>\n" );
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document.write( "thanks in advance for helping me  </SPAN>\n" );
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document.write( "With compass bearings, the angle 000° points directly north. As the angle increases, you turn toward the east direction. This means the angle increases as you turn clockwise as this diagram indicates\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" width=\"50%\" src=\"https://i.stack.imgur.com/tlVq5.gif\">\r<BR>\n" );
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document.write( "Image source: <BR>\n" );
document.write( "https://gis.stackexchange.com/questions/239374/determining-spatial-sort-order-with-arcpy\r<BR>\n" );
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document.write( "The bearing 240 degrees is located in the southwest quadrant, while the bearing 025 degrees is in the northeast quadrant.\r<BR>\n" );
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document.write( "When it comes to angles in trig, convention has 0 degrees pointing directly east. <BR>\n" );
document.write( "Also, the angle increases as you move counterclockwise.<BR>\n" );
document.write( "So we'll have to make a conversion. \r<BR>\n" );
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document.write( "Note how the bearing 240 degrees is 30 degrees below the west direction (270). <BR>\n" );
document.write( "For angles in trig, the west direction is 180 degrees.<BR>\n" );
document.write( "This means we add on 30 to 180 to get 30+180 = 210\r<BR>\n" );
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document.write( "So the bearing 240 degrees is the same as the angle 210 degrees when converting to the trig equivalent angle.\r<BR>\n" );
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document.write( "Similarly, the bearing 025 degrees converts to 65 degrees\r<BR>\n" );
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document.write( "This diagram visually summarizes what I mean<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" width = \"80%\" src = \"https://i.imgur.com/oOgRTpa.png\"><BR>\n" );
document.write( "note: The diagram above showing the plane's vector is not accounting for the wind.\r<BR>\n" );
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document.write( "We have the bearing 025° be the same as the angle 65°, when we start 0 aimed at the east direction (instead of north)\r<BR>\n" );
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document.write( "The bearing 240° is the same as the angle 210°\r<BR>\n" );
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document.write( "The bearings are in red while the converted angle measures are in blue\r<BR>\n" );
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document.write( "I broke up the two vectors so they each had their own coordinate system. Though we'll combine them later as explained below.\r<BR>\n" );
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document.write( "The vector for the aircraft, without accounting for the wind, is<BR>\n" );
document.write( "< x, y > = < r*cos(theta), r*sin(theta) ><BR>\n" );
document.write( "< x, y > = < 200*cos(210), 200*sin(210) ><BR>\n" );
document.write( "p = < 200*cos(210), 200*sin(210) >\r<BR>\n" );
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document.write( "Where r = 200 is the plane's speed without the wind affecting it\r<BR>\n" );
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document.write( "The wind's vector is<BR>\n" );
document.write( "< x, y > = < r*cos(theta), r*sin(theta) ><BR>\n" );
document.write( "< x, y > = < 60*cos(65), 60*sin(65) ><BR>\n" );
document.write( "w = < 60*cos(65), 60*sin(65) >\r<BR>\n" );
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document.write( "Now we add the two vectors:\r<BR>\n" );
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document.write( "p+w = < 200*cos(210), 200*sin(210)>+<60*cos(65), 60*sin(65) ><BR>\n" );
document.write( "p+w = < 200*cos(210)+60*cos(65), 60*sin(65)+200*sin(210) ><BR>\n" );
document.write( "p+w = < -147.847985052446, -45.621532777801 ><BR>\n" );
document.write( "f = < -147.847985052446, -45.621532777801 ><BR>\n" );
document.write( "Make sure your calculator is in degree mode.\r<BR>\n" );
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document.write( "f represents the final vector, which is where the plane is aimed with the wind accounted for\r<BR>\n" );
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document.write( "Let f = < a,b >\r<BR>\n" );
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document.write( "We need to find the direction and magnitude for vector f\r<BR>\n" );
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document.write( "Recall the magnitude is simply the length of the vector. <BR>\n" );
document.write( "We use the pythagorean theorem to see that<BR>\n" );
document.write( "|f| = sqrt(a^2+b^2)<BR>\n" );
document.write( "|f| = sqrt((-147.847985052446)^2+(-45.621532777801)^2)<BR>\n" );
document.write( "|f| = 154.726697557546<BR>\n" );
document.write( "|f| = 154.73\r<BR>\n" );
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document.write( "After accounting for the wind, the plane's speed is roughly 154.73 mph.\r<BR>\n" );
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document.write( "Now find the direction angle theta<BR>\n" );
document.write( "theta = arctan(b/a)<BR>\n" );
document.write( "theta = arctan(-45.621532777801/(-147.847985052446))<BR>\n" );
document.write( "theta = 17.1486847735892<BR>\n" );
document.write( "theta = 17.15\r<BR>\n" );
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document.write( "This theta value points in quadrant 1 (northeast quadrant), but we want to aim for quadrant 3 (southwest quadrant).<BR>\n" );
document.write( "This is because both a,b are negative making the point (a, b) be in quadrant 3.<BR>\n" );
document.write( "So we need to add 180 to this result\r<BR>\n" );
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document.write( "theta+180 = 17.15+180 = 197.15\r<BR>\n" );
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document.write( "After accounting for the wind, the plane's direction is roughly 197.15° (which is in the southwest quadrant)\r<BR>\n" );
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document.write( "<font color=red>Answer: d. The speed is 154.73 miles/hour, and the direction angle is 197.15°</font></font> </SPAN>\n" );
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