2004 HSC please clarify :) (1 Viewer)

[crammy90]

2004 HSC question 6 b ii
it says "suppose that it is given that AY bisects angle BAC.
in the answers they have assumed this bisection is into two equal parts
is bisection in maths assumed to be into two equal parts?
thanks

and question 5 b iv no idea lol

 

[tacogym27101990]

2004 HSC question 6 b ii
it says "suppose that it is given that AY bisects angle BAC.
in the answers they have assumed this bisection is into two equal parts
is bisection in maths assumed to be into two equal parts?
thanks

and question 5 b iv no idea lol

bisection means its cut into 2 equal parts
its disection that your thinking

and ill just give you a hint for 5 , speed is greatest when accel is 0

 

[Pwnage101]

by definition of bisection, u can assume 2 equalo parts....

 

[crammy90]

ahk thanks heaps for that clarification.

for get acc DV/Dt : -12cos2t (/-12)

cos2t = 0
now lost...
​

in the ansers they then do
2t = pi/2, 3pi/4
t = pi/4, 3pi/4 -->i see here they have multiplied by 1/2 to get t on its own. where do the original numbers pi/2, 3pi/4 come from?
thanks heaps for fast replies
​

 

[vds700]

ahk thanks heaps for that clarification.

for get acc DV/Dt : -12cos2t (/-12)

cos2t = 0
now lost...
​

in the ansers they then do
2t = pi/2, 3pi/4
t = pi/4, 3pi/4 -->i see here they have multiplied by 1/2 to get t on its own. where do the original numbers pi/2, 3pi/4 come from?
thanks heaps for fast replies
​

they have simply solved the trigonometric equation, cos2t = 0
2t = cos<meta http-equiv="Content-Type" content="text/html; charset=utf-8"><meta name="ProgId" content="Word.Document"><meta name="Generator" content="Microsoft Word 11"><meta name="Originator" content="Microsoft Word 11"><link rel="File-List" href="file:///C:%5CUsers%5CAndrew%5CAppData%5CLocal%5CTemp%5Cmsohtml1%5C01%5Cclip_filelist.xml"><!--[if gte mso 9]><xml> <w:WordDocument> <w:View>Normal</w:View> <w:Zoom>0</w:Zoom> <wunctuationKerning/> <w:ValidateAgainstSchemas/> <w:SaveIfXMLInvalid>false</w:SaveIfXMLInvalid> <w:IgnoreMixedContent>false</w:IgnoreMixedContent> <w:AlwaysShowPlaceholderText>false</w:AlwaysShowPlaceholderText> <w:Compatibility> <w:BreakWrappedTables/> <w:SnapToGridInCell/> <w:WrapTextWithPunct/> <w:UseAsianBreakRules/> <wontGrowAutofit/> </w:Compatibility> <w:BrowserLevel>MicrosoftInternetExplorer4</w:BrowserLevel> </w:WordDocument> </xml><![endif]--><!--[if gte mso 9]><xml> <w:LatentStyles DefLockedState="false" LatentStyleCount="156"> </w:LatentStyles> </xml><![endif]--><style> <!-- /* Style Definitions */ p.MsoNormal, li.MsoNormal, div.MsoNormal {mso-style-parent:""; margin:0cm; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:"Times New Roman"; mso-fareast-font-family:"Times New Roman";} @page Section1 {size:612.0pt 792.0pt; margin:72.0pt 90.0pt 72.0pt 90.0pt; mso-header-margin:36.0pt; mso-footer-margin:36.0pt; mso-paper-source:0;} div.Section1 {page:Section1;} --> </style><!--[if gte mso 10]> <style> /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:10.0pt; font-family:"Times New Roman"; mso-ansi-language:#0400; mso-fareast-language:#0400; mso-bidi-language:#0400;} </style> <![endif]-->^-10 put this ion your calc and you'll get 90 degrees.

as cos2t is positive, the other solution must be in the 4th quadrant (Remember ASTC)

therefore 270 degrees (or 3pi/4) is a solution,

2t = pi/2, 3pi/2
t = pi/4, 3pi/4
Maybe you need to revise angles of any magnitude and the unit circle.