Quick Question, Help (1 Viewer)

[adonis1]

I got a math question, i drew it up in paint, just download the attachement and show me how to do it if u can.

Cheers

 

[FinalFantasy]

angle CBA=65
angle CBJ=60
therefore angle JBA=5 and angle BJA=150
10\sin5=BJ\sin25
BJ=sin25*10\sin5

sin30=BC\BJ
BC=BJsin30

 

[adonis1]

ty, im still confused because i think you used further trigonemetry and we still havent learnt that crap in 2 unit so far.

 

[FinalFantasy]

i didn't use any "further trignometry"...
angle CBA=65 (180-25-90)
angle CBJ=60 (180-30-90)
therefore angle JBA=5(65-60) and angle BJA=150(180-30)
den i use a\sina=b\sinb=c\sinc
the last one i used sin x=opposite over hypotenuse

 

[nick1048]

yer he's right man, sit down with the diagram and draw up the angles he's stated, it makes sense when u can visualise whats happening.

 

[adonis1]

Cheers, but the thing is the stuff u used Sina/c etc all that stuff, we havent learnt that, I know of it like in my maths book ive seen it but we still havent learnt it, and im going through my maths book to see if i wrote it down in cased i missed it out, but it isnt there. Thanks for the help mate, ill try and revise this topic to see if later ill understand it.

Cheers.

 

[nick1048]

u dont noe the sina/a rule... like for real? That's like the core of the 2u topic, what do u noe about trig???

 

[adonis1]

seriously we havent learnt that, like wat our class knows is sohcahtoa, inverse ratios, the two triangles to find the exact values of sin tan and cos 60, 30, 45 etc. Umm what else, oh ye, bearings, trig ratios, angles of magnitude and trig equations like

cot30 +tan 60
__________

tan45 + cot 30

stuff like that, but we havent done the sin rule and cos rule stuff what u talking about, but i am aware of what u are talking about, i just dont know how to do it cause we havent learnt it yet.

 

[mouldy_bread]

i didn't use any "further trignometry"...
angle CBA=65 (180-25-90)
angle CBJ=60 (180-30-90)
therefore angle JBA=5(65-60) and angle BJA=150(180-30)
den i use a\sina=b\sinb=c\sinc
the last one i used sin x=opposite over hypotenuse

I always hated trigonometry. What is sin, cos and tan anyway? They arent real numbers... like pi. So annoying

 

[Trev]

Aren't they ratios?
Correct me if im wrong, which I think I am!

 

[Estel]

If I may offer my humble Yr 7 opinion: trigonometric ratios are actually exponential functions, but the link is far beyond the 2 unit course. If you accept squaring, and cubing etc there's no reason why you shouldn't accept trig.

 

[Trev]

Never woulda known Estel.
Curious - what marks are you aiming for in extension 1 and 2 mathematics?

 

[Estel]

Haven't thought much about that to be honest.
100 in 3U is definitely the aim, as for 4u I'm not sure yet. Maths is getting quite dry for me though. =/

 

[FinalFantasy]

ooer Estel i thought u were class of 04

 

[Trev]

Haven't thought much about that to be honest.
100 in 3U is definitely the aim, as for 4u I'm not sure yet. Maths is getting quite dry for me though. =/

My god! lol, probably about 98-99 in 4u then ey?
What school do you go to?

 

[mouldy_bread]

If I may offer my humble Yr 7 opinion: trigonometric ratios are actually exponential functions, but the link is far beyond the 2 unit course. If you accept squaring, and cubing etc there's no reason why you shouldn't accept trig.

hmmmm.... definately feeling in over my head at the moment.

 

[Estel]

Trev... wait for it... a regional backwater called Sefton!
Don't laugh... =p

Must beat Hong...

In all seriousness I doubt I'd get a 99 in 4U. Don't really want it enough, and in all truth probably not talented enough for it to come on a silver platter to me.

 

[mouldy_bread]

if its any consolation I think theres a strong chance it could be. I'm hoping for a mediocre maths mark so I can concentrate on my better subjects.

 

[Slidey]

Never woulda known Estel.
Curious - what marks are you aiming for in extension 1 and 2 mathematics?

He is correct. I forget how to derive it, but it can be seen from Euler's formula:
e^(@.i)=cos@+i.sin@

Hmm. Maybe...
(1) e^(@.i)=cos@+i.sin@
(2) e^(-@.i)=cos@-i.sin@
[(1)+(2)]/2:
[e^(@i)+e^(-@i)]/2=cos@
[(1)-(2)]/(2i):
[e^(@i)-e^(-@i)]/(2i)=sin@

That is:
cos(x)=(e^ix+e^-ix)/2
sin(x)=(e^ix-e^-ix)/2i

You've also got hyperbolic trig. Normal trig is based on the unit circle. Hyperbolic trig (sinh(@), tanh(@), arccosh(@), et cetera) is based on the unit hyperbola.

And just like for circle trig, sin^2(@)+cos^2(@)=1, for hyperbolic trig:
sinh^2(@)-cosh^2(@)=1

Furthermore, hyperbolic trig ratios can ALSO be expressed in exponential form. Don't ask me how to derive that one, though.

Although I do remember that cos(z)=cosh(iz), so it's probable that it is:
cosh(x)=(e^x+e^-x)/2

 

[Trev]

Trev... wait for it... a regional backwater called Sefton!
Don't laugh... =p

Must beat Hong...

In all seriousness I doubt I'd get a 99 in 4U. Don't really want it enough, and in all truth probably not talented enough for it to come on a silver platter to me.

I have no clue where Sefton is, although i'm already certain it would beat my school!
Hong, the guy that came 2nd in extension 2? I guess you will have to come first then, maybe it will be another year where 'the best' screw up, and you end up coming first!