Part A Part B Part C = document.getElementById(\'wrapper_1000321_problemView1\')

Question

Part A Part B Part C = document.getElementById('wrapper_1000321_problemView1').innerHTML = MyCT_FL_RunContent('codebase','http://download.macromedia.com/pub/shockwave/cabs/flash/swflash.cab#version=8,0,34,0','id','problemView1','name','problemView1','tabindex','0','width','375','height','60','src','/assets/flash/eq_entry.build1009161005','movie','/assets/flash/eq_entry.build1009161005','quality','high','align','middle','base','/problemAsset/1000321/21/','FlashVars','lcId=LC_problemView1&objectID=problemView1&helpURL=http://www.masteringsupport.com/product_help/help_students/Content/TopicsStudent/enteringmath.htm&shortcutsURL=http://www.masteringsupport.com/palette_shortcuts/math_shortcuts.htm&maxWidth=375&maxHeight=480&bordercolor=ffffff&toolbarloc=/assets/flash/eq_entry/science/&xl=1&engineering=0&setResponseValue=','seamlessTabbing','true','allowFullScreen','false', 'allowScriptAccess', 'sameDomain', 'pluginspage','http://www.macromedia.com/go/getflashplayer', 'wmode', 'window');

Explanation / Answer

from figure vector A = 4 i vector C = -2 j A+B+C = -1.6i So, vector B = -1.6 i -A- C =-1.6 i -4i+2j = -5.6 i + 2j vector B = -5.6 ,2 (b). magnitude of vector B = sqrt [ -5.6^ 2+ 2 ^ 2] = 5.946 units (c). let B makes an angle theta with negative x-axis then tan(theta) =2 /5.6 theta = 19.65 degrees

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