# Finding exact values

• Jun 14th 2011, 09:19 AM
NecroWinter
Finding exact values
http://i250.photobucket.com/albums/g...troph/trig.png

I dont really know what to do to solve this problem. It wants me to find exact values of each part labeled with a variable.

I have the solution manual, but it doesnt go into any kind of depth, and sort of assumes that I know what its talking about, or how it got certain things. For example, it starts off by looking at variable y and states = y=(1/2)(9)
I dont know where the 1/2 comes from.

Can anyone give me some information? Usually a section in my book has an example to go along with it, this one doesnt. Please explain how you know something to be true in the process.
• Jun 14th 2011, 09:39 AM
abhishekkgp
Re: Finding exact values
Quote:

Originally Posted by NecroWinter
http://i250.photobucket.com/albums/g...troph/trig.png

I dont really know what to do to solve this problem. It wants me to find exact values of each part labeled with a variable.

I have the solution manual, but it doesnt go into any kind of depth, and sort of assumes that I know what its talking about, or how it got certain things. For example, it starts off by looking at variable y and states = y=(1/2)(9)
I dont know where the 1/2 comes from.

Can anyone give me some information? Usually a section in my book has an example to go along with it, this one doesnt. Please explain how you know something to be true in the process.

i guess side 'y' is perpendicular to side 'x' and 'z'. if this is assumed then its just plain trig.
• Jun 14th 2011, 10:07 AM
NecroWinter
Re: Finding exact values
Quote:

Originally Posted by abhishekkgp
i guess side 'y' is perpendicular to side 'x' and 'z'. if this is assumed then its just plain trig.

could you be a little more specific?
• Jun 14th 2011, 10:16 AM
Plato
Re: Finding exact values
Quote:

Originally Posted by NecroWinter
could you be a little more specific?

$\displaystyle \tan \left( {30^o } \right) = \frac{y}{x}\;\& \,\tan \left( {60^o } \right) = \frac{y}{z}$

Just keep that up to get all the relations so that you can find $\displaystyle x,~y,~z,~\&~w$.
• Jun 14th 2011, 11:39 AM
NecroWinter
Re: Finding exact values
The way my book explains it is as follows:

(1/2)(9) & x=y(3)^(1/2)

then

y=z(3)^(1/2), so z = (9/2)/(3)^(1/2) = 3(3)^(1/2)

and

w=2z, so w = 2((3(3)^(1/2)/2) = 3(3)^(1/2)

I kind of understand the first line by simply pretending like the triangle is cut in half, and was originally an equilateral triangle, can anyone explain whats going on from there on?
• Jun 14th 2011, 12:23 PM
Also sprach Zarathustra
Re: Finding exact values
y=3

x^2+y^2=9^2 ==> x^2=81-9=72 ==>x=sqrt(72)

y^2=x*z (geometric mean)
• Jun 14th 2011, 12:35 PM
HallsofIvy
Re: Finding exact values
Yes, you are cutting an equilateral triangle into two parts. Let's say each side of the equilateral triangle has length 2. Then the side opposite the 30 degree angle has length 1. From the Pythagorean theorem, if x is the length of the side opposite the 60 degree angle, $\displaystyle x^2+ 1^2= 2^2$ so $\displaystyle x^2= 4-1= 3$ and $\displaystyle x= \sqrt{3}$.

That is, your right triangle has legs of length 1 and $\displaystyle \sqrt{3}$ and hypotenuse of length 2. Now use "sine= opposite side/hypotenuse", "cosine= near side/hypotenuse", etc..
• Jun 14th 2011, 03:41 PM
skeeter
Re: Finding exact values
Quote:

Originally Posted by NecroWinter
http://i250.photobucket.com/albums/g...troph/trig.png

I dont really know what to do to solve this problem. It wants me to find exact values of each part labeled with a variable.

review the side length ratio relationships for 30-60-90 triangles ...

http://img818.imageshack.us/img818/1029/306090.jpg