Help with FST exam

[gamer2]

I need help understanding some things

such as
- finding Arc length in radians
-finding a Missing log such as (log c=1.03, log b-.98575, log c=b/c)
-Finding Missing angles in triangles using Laws of Cos/Sines

Please help

[Gamelver]

heh, I'm studying all this stuff for an exam tomorrow ...

anyway, I haven't...actually....er...looked over any of that stuff ...but I know that the arc length equals the radius times theta in radians (theta being the value of the angle making the arc...whatever the hell that's called). The Law of Cosines/Sines are just some equations...know them and plug values in, really. Oh, and for that arc length thing, make sure you can convert degrees into radians...just remember, 360 degrees equals 2*pi radians ...so it's something like (pi/180) times the degrees equals radians....and perhaps (180/pi) times radians equals degrees....

A log is like...well.....log(100)=2

when you have a log of something, it's like saying the log of ten (unless a different subscript number is next to log) is raised to the second power to get 100 (hence, the log(100)=2).

so...the log5(125)=3 because 5^3=125. does that make sense ?

Edit: LINK!!!

[im_an_alien]

Finding Missing angles in triangles using Laws of Cos/Sines

You mean like when you have a right triangle and you have 2 side lengths and you have to find an angle measure?

[Krepticor]

here is a few things googled up

here


try these and review your basic logirithm relationships

ie. log(a) + log(b) = log(a*b)

log(a/b) = log(a)-log(b)

also remeber that logirithms are the opposite of exponents

ie. log(x)=y means that x to the tenth power is y (it is assumed in calculus that anytime logis not stated it is at the base ten, just as ln is the natural log which is at the base e)

[Gamelver]

also, a*log(b)=log(b^a), I believe ...

[Skyone]

Jeez - can't you just do sin x (a/b)?

[xelion]

I always hated logs, log of the number over log of the base. log N/log B, and B^p=N Base to the power equals number. Not sure if this is what you are wanting, btu it's what I've learned so far, take for example 2.31^X=100 (numbers are made up and probably won't work) The x is what you are trying to get, so you take log of the number(100) by log of the base(2.31) to find the exponent x, log(100)/log(2.31)

Logs are usually used like that to find exponents and such, basic properties of logs, log(a)-log(b) = log(a/b) or log(a)+log(b)=log(a*b) and there was one, like the base of 10 is used in logs, log10(a) is the same as log(a) but when a number is given, such as log2(a) it's not the same, so you have to change it back to the B^p=N form, 2 is the base and a is the number so.. 2^p=a, and use the log(N)/log(B) to find the power, P. And umm, I think like if its log2(2)^x then the answer is 2. If the log base is the same as the number, then the power x is the same thing.

I'm not in trig or anything, but are you talking about SOHCAHTOA? Used to find the missing angles in right triangles?

|\
| \ Hypotenuse
|__\
either of the other sides are oppisite depending on which angle you are using.

sin(x)=opp/hyp
cin(x)=angle/hyp
tan(x)=opp/angle

I'm not sure how much of that is right, been so long. As for the other thing, I don't remember that part.

[Gamelver]

law of sines: sin(alpha)/a=sin(beta)/b=sin(gamma)/c

alpha is an angle, and a is the side opposite that angle...same for the other parts of the equation.

the law of cosines is pretty complicated, but you can just plug in values for the angles and sides and figure it out . Here it is: c2 = a2 + b2 – 2ab cos C

or: a2 = b2 + c2 – 2bc cos A

or: b2 = a2 + c2 – 2ac cos B

yes, I copied those ...

[gamer2]

I know how to do right triangles, just use the old indian SOHCAHTOA (Sin=Opoite/Hypotenuse, Cos=Adj/Hypotenuse, Tan=Oposite/Adj)

I need help with other triangles