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Mathematics 9 Online
OpenStudy (saifoo.khan):

Q3 http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s08_qp_3.pdf part 1.

OpenStudy (anonymous):

are you trying to get into Cambridge?

OpenStudy (anonymous):

Which Question ?

OpenStudy (anonymous):

Q3

OpenStudy (saifoo.khan):

@tomo : I already study in a Cambridge system.

OpenStudy (anonymous):

nice

OpenStudy (anonymous):

Sorry @saifoo.khan I don't know the Solution

OpenStudy (saifoo.khan):

@tomo thanks @Tribute2SarabjitSingh np @hartnn C'mon man!!

OpenStudy (anonymous):

the absolute value one?

OpenStudy (saifoo.khan):

Q3

OpenStudy (saifoo.khan):

Is it that hard? @satellite73

OpenStudy (anonymous):

2(2r+xr)=8a

OpenStudy (anonymous):

r=a/sinx

OpenStudy (saifoo.khan):

Or do you mean 2(2x+xr) = 8a ?

OpenStudy (anonymous):

The perimeter of the sector AMN is equal to half the perimeter of the rectangle.

OpenStudy (saifoo.khan):

Ah! RIGHT! I made a mistake.

OpenStudy (anonymous):

ohh sorry tehre 2r instead of 2x

OpenStudy (saifoo.khan):

Whats next?

OpenStudy (anonymous):

substitute 2(2r+xr)=8a 4r+2xr=8rsinx

OpenStudy (anonymous):

divide by 2r

OpenStudy (anonymous):

2+x=4sinx

OpenStudy (anonymous):

sinx = 1 4 (2+x).

OpenStudy (saifoo.khan):

Awesome! What about part (ii) ?

OpenStudy (saifoo.khan):

In part (ii), we just have to insert 0.8?

OpenStudy (saifoo.khan):

Wait. i got it.

OpenStudy (saifoo.khan):

Thanks @SerikMB

OpenStudy (anonymous):

how did u solve it?

OpenStudy (anonymous):

just substitute four times?

OpenStudy (saifoo.khan):

first insert 0.8, then insert the answer back, then insert the answer again. We will get a series of answers. 0.7754 0.7668 0.7638 0.7628 Since we needed the answer correct to 2dp, so our answer will be 0.76

OpenStudy (anonymous):

thx u too

OpenStudy (saifoo.khan):

np

OpenStudy (saifoo.khan):

Wanna try q4?

OpenStudy (anonymous):

sorry, i didn't notice ur reply, well it's very easy but quite tedious \[\tan(30+\theta)=2\tan(60-\theta)\]

OpenStudy (anonymous):

\[\frac{ \tan30-\tan \theta }{ 1-\tan30*\tan \theta }=2*\frac{ \tan60-\tan \theta }{ 1+\tan60*\tan \theta }\]

OpenStudy (anonymous):

tan60=square root 3 tan30=square root 3 over just 3

OpenStudy (anonymous):

take tan(theta) as x to make calculation easier

OpenStudy (anonymous):

\[\frac{ \frac{ \sqrt{3} }{ 3 } +x}{ 1-\frac{ \sqrt{3} }{ 3 }*x }=2*\frac{ \sqrt{3}-x }{ 1+\sqrt{3}*x }\]

OpenStudy (anonymous):

by the calculation you will get \[\sqrt{3}*x ^{2}+18x-5\sqrt{3}=0\]

OpenStudy (anonymous):

divide by \[\sqrt{3}\]

OpenStudy (anonymous):

and you will get the answer

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