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Question 9b i) (1 Viewer)

kynanld

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Apr 1, 2005
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I had few problems up untill this question, and it totally threw me how do you do it?
 

jemmi

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well i dunno if i got this right but here's my working...

SinΦ = BD/6

BD = 6SinΦ

then

SinΦ = ED/ 6SinΦ

ED = 6Sin^2Φ

then

SinΦ = EF/ 6Sin^2Φ

EF = 6Sin^3Φ
 

kynanld

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Apr 1, 2005
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Ohh man thx heeps for that. I immediatly decided the sin and cos rule wern't gonna help because for some bad reason I assumed I was going to find a constant for DB.
Yea well done looks right to me 2.
 

Haku

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i think u need to say that the triangles are similar. not sure if proving it is required... didn't have time, so just same they were similar
 

chin music

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This question really annoyed me. Apparantley angle DBE is theta. too bad i only realised that after the exam
 

Jago

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yep i only saw it because i put in the ^ thing to see the parallel lines
 

MarsBarz

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Or simply,
angle DAB = theta
angle DBA = 90 - theta
angle DBE = 90- (90- theta) = theta
Etc...
 

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