help with a simple question (1 Viewer)

[FOB all the way]

this question is causing me issues, its from the 97 paper
the point p lies on the circumference of a semicircle of radius r and diameter AB. the point C lies on AB and PC is perpendicular to AB. THE arc AP subtends an angle @ (theta) at the centre O and the length of the arc AP is twice the length of PC

i) Show that 2sin@=@

ii) Taking @=1.8 as an approximation for the solution to the equation 2sin@=@ between pi/2 and pi, use one application of Newtons method to give a better approximation

There is a diagram provided but i dont have a scanner but here is a link to the question
http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/97MAT3U.PDF
thanks in advance

 

[friction]

Draw a picture it always helps. I cant be bothered to read the question.

 

[QuLiT]

2PC = arc AP
2rsin@=r@
2sin@=@

View attachment 17208

 

[FOB all the way]

the diagram is provided so thats not the issue, its more the proof, but sorry to bother you, why are you browsing this site if you could not be bothered to help anyone?

 

[friction]

l = r(theta) (length or arc rule)
l = 2PC (given)
2PC = r (theta)

PC = rsin (theta) (sin theta = opp/adj)

2 sin (theta) = theta

 

[friction]

Lol i decided to do it after i got something to eat. It had already been done .

Needed a break just did an english paper.

second part no trouble im guessing ??

 

[Marinatos]

For the Newton's approximation part just use the formula b = a - f(a)/f'(a)

 

[friction]

second part is is second guess = (first guess) - ( f(theta)/ f'(theta))

 

[FOB all the way]

thanks for your help
yer second part no trouble
good luck every one for the HSC
i hope extension 1 is good to us all

 

[friction]

ext 1 isnt the big problem.

 

[FOB all the way]

yer still i am hoping to maximise my marks there to bost my uai cause english is going to bring it down