Question 9b i) (1 Viewer)

[kynanld]

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

 

[jemmi]

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

 

[MajinR]

looks about right

 

[kynanld]

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]

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

 

[Ravo_Kate]

-

 

[chin music]

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

 

[Jago]

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

 

[MarsBarz]

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