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PostPosted: Fri Mar 10, 2006 12:21 am 
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Math wizes, please help me out. For a jig I'm making, I need to calculate the dimensions of a triangle. It's a right triangle. I had to plumb the depths of my high school math memory just to remember the word "hypotenuse". I know the length of the hypotenuse, and the angle of the more acute of the two angles other then the 90 degree one. It seems like that's all I should need to know to calculate the length of the two legs and the third angle. Actually, I think I've already figured out that the three angles of a triangle always add up to 180 deg, right? So, if the one acute angle is 25 degrees, the other must be 65. Right? But I still need to know the length of the legs. What's the formula for that?

Thanks!

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PostPosted: Fri Mar 10, 2006 12:32 am 
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Todd,

If the length of your hypotenuse is h then the lengths of the other 2 sides are given by:

h x sin(25) and
h x cos(25)

or alternatively:

h x sin(65) and
h x cos(65)

Should give the same results.

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PostPosted: Fri Mar 10, 2006 12:40 am 
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Brazilian Rosewood
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Ahhhh, the law of sines reers it's ugly head. Here's an expample I found. I'm not sure where your numbers relate so you can plug them in and use this formula.

       

                   B
                  /^\<this angle measures 105 degrees
       15 km> /    \
                /        ;\   
               /        ; \
              /        ;   ;    \
        A /_ _ _ _ _ _ _ _ _\C
this is 45 degrees     this is, consequently, 30 degrees



You could use the Law of Sines, which states

        sin(A)   sin(B)   sin(C)
        ------ = ------ = ------ ,
              a        ; b        c

where A, B, C are angles and a, b, c are the lengths of the sides they
subtend (are opposite to). So side AB is "c" in the above equation.
sin(A) and sin(C) are easy to find; they are 1/sqrt(2) and 1/2,
respectively.

sin(B) = sin(105) = sin(45+60) = sin(45)cos(60)+cos(45)sin(60) =
(1/sqrt(2))*(sqrt(3)/2)+(1/sqrt(2))*(1/2) = (1+sqrt(3))/(2*sqrt(2)).

So we have

           1/sqrt(2)   (1+sqrt(3))/(2*sqrt(2)) &nbs p; 1/2
        --------- = ----------------------- = --- ,
             a                     b                   15

so b = 15(1+sqrt(3))/sqrt(2) = AC, and a = 15*sqrt(2) = BC, which
agrees with our previous results.

In general, area considerations are a poor way of obtaining relations
between angles and sides, because they are often very complicated and
often come in a form that requires knowing the lengths of more than
one side. If you know a lot of angles, a better approach is to think
of the Law of Sines or the Law of Cosines
(c^2 = a^2+b^2-2*a*b*cos(C)).

In general, given a side and two angles,
you must use the Law of Sines to find the other lengths.

Pwoolson38786.3740740741


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PostPosted: Fri Mar 10, 2006 12:41 am 
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Koa
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The sine of an angle is the ratio of the hypotenuse over the opposite side. So, if your acute angel is 25 degrees, the opposite side is .412618262... as long as the hypotenuse. If the hypotenuse is 10 inches the short side of the triangle is about 4.2 inches.

The cosine is the ratio of the hypotenuse over the adjacent side. For a 25 degree angle that ration is: .906307787..
Or, for a 10 inch hypotenuse about 9.0 inches.

You are right that the sum of the angles in any triangle is 180 degrees.


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PostPosted: Fri Mar 10, 2006 1:11 am 
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Heh guys please stop this! I come to the forum to escape all the maths (anyway I've got a Cray to do the sums).

Colin

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PostPosted: Fri Mar 10, 2006 1:48 am 
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Oh, man, I feel just like I felt in 10th grade geometry - it was my worst subject and I never took another math class after that. That was 27 years ago. Sines and cosines! Oy! My head is spinning. I'm sorry, Paul, I can't follow your equations. You might as well be writing in Chinese as far as I'm concerned. Dave, I like the simplicity of what you stated, but I don't know how to calculate the sines and cosines. Also, let me make sure I follow...

h x cos(25)

This means h MULTIPLIED BY the cosine MULTIPLIED BY 25, right?

Now, to calculate the sine and cosine...

Mike says, "The sine of an angle is the ratio of the hypotenuse over the opposite side."

Do you mean the ratio of the LENGTH of the hypotenuse over the LENGTH of the opposite side, that is, opposite the angle in question?

At any rate, since I don't know the length of the opposite side (that's what I want to find out), I'm not seeing how this gives me the formula into which I can plug the numbers I do know.

I see that you've given me a number for a 25 degree angle: .412618262. Trusting that you are correct, I could use that number to calculate the length of my side. But I don't see how you arrived at that number, and I'd like to know, so I could do my own calculations with other angles.

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PostPosted: Fri Mar 10, 2006 2:01 am 
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Todd,

Mike has done what I was doing more precicesly for you. It is h multiplied by the sine of 25 degrees, or h multiplied by 0.412618262; and h multiplied by the cosine of 25 degrees or h multiplied by 0.906307787.

If stuck you can find the functions in Excel.

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". . . the one thing a machine just can't do is give you character and personalities and sometimes that comes with flaws, but it always comes with humanity" Monty Don talking about hand weaving, "Mastercrafts", Weaving, BBC March 2010


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PostPosted: Fri Mar 10, 2006 2:08 am 
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It's h MULTIPLIED by the COSINE OF 25...

Windows comes with a calculator (if you're using windows), go to Programs, Accessories, Calculator. Go to View and change it to Scientific. Make sure the little radio button is set for Degrees, not Radians or Grads.

Type in 25. Then hit COS. The result is cos(25). Then multiply that by h.

Or, find a free cad program and just draw it out and measure it if you don't trust your math.

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PostPosted: Fri Mar 10, 2006 2:16 am 
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Okay, I just discovered that my calculator has a "sin" button and a "cos" button. Never used those buttons before. Oddly, though, when I enter 25 then push the sin button, it's giving me .4226183... allright, I see that's just one digit off from the number you guys are stating... probably just a typo, eh, Mike?

So, am I to understand that if you want to know the sine or cosine of an angle, you just grab a calculator? Is the equation used to calculate it on paper incredibly complicated?


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PostPosted: Fri Mar 10, 2006 2:29 am 
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As a general rule, you don't calculate the sine or the cosine of an angle. You look them up. They are universal constants. Your friendly calculator can do that for you. The sine of 25 degrees is always .422618262.

Long ago it was discovered that the ratio of the length of the hypotenuse to the length of the sides of a right triangle that had any particular angle would never change regardless of the size of the triangle provided that the angle didn't change. These ratios were given the names sine, cosine and tangent.
sine = hypotenuse / opposite side
cosine = hypotenuse / adjacent side
tangent = opposite side /adjacent side

For a right triangle the side opposite the 25 degree angle is the that shortest side.


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PostPosted: Fri Mar 10, 2006 2:33 am 
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[QUOTE=Todd Rose] Okay, I just discovered that my calculator has a "sin" button and a "cos" button. Never used those buttons before. Oddly, though, when I enter 25 then push the sin button, it's giving me .4226183... allright, I see that's just one digit off from the number you guys are stating... probably just a typo, eh, Mike?

So, am I to understand that if you want to know the sine or cosine of an angle, you just grab a calculator? Is the equation used to calculate it on paper incredibly complicated?

[/QUOTE]

Yes, it is a typo. The number my caclulator give is:
0.422618262

If you want to calculate it on paper, it is quite a bit more complicated than what most people are willing to put up with. Since it never changes, people just look it up.



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PostPosted: Fri Mar 10, 2006 2:36 am 
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Interesting...I really liked Trigonometry and Geometry....
...specifically all this right triangle Trig stuff.
Don Williams38788.4403356481

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PostPosted: Fri Mar 10, 2006 2:38 am 
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Listen to Dave. You don't need the law of sines or law of cosines if there is a right angle.

Ron

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PostPosted: Fri Mar 10, 2006 3:12 am 
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Allright, guys, thanks! You have answered my questions and I'm on my way, calculator in hand! Cool!

Funny thing is, aside from my first reaction at seeing Paul's post (no offense, Paul - thanks for trying!), I actually enjoyed this discussion! I kinda got this, "hey, trigonometry is cool!" feeling, rather than the dread and hopelessness I always associated with my memories of 10th grade geometry (I was a straight A student until I FAILED the final exam in that class. Ouch.) Maybe math ain't so bad after all...

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PostPosted: Fri Mar 10, 2006 5:14 am 
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Brazilian Rosewood
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Don't worry too much about maths Todd, i was in the same boat when younger, alweays had a hard time with it maybe because i'm a visual person and couldn't see the purpose of it all unless the measurements were all written down!


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PostPosted: Sat Mar 11, 2006 2:01 pm 
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So there are these three indian (native american) women sitting around a camp and two are talking about thier sons.

One says "look ther at my lodge the buffalo skin hanging there. My son killed it and skinned it and gave it to me. My son is the greatest hunter".

The next woman says " look there at my lodge, the bear skin hanging there. My son killed it and skinned it and gave it to me. All with just a knife! My son is the greatest hunter".

The third, quiet until now stands and says "look there at my lodge, the hippopotamus hide hanging there. I killed it and I skinned it and I hung it on my own lodge. I am just as great a hunter as both of your sons".

The moral of the story?


The squaw of the hippopotamus hide is equal to the sons of the squaws of the other two hides...

Math geek joke.


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PostPosted: Sat Mar 11, 2006 4:25 pm 
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Brilliant


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PostPosted: Sat Mar 11, 2006 6:28 pm 
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[QUOTE=Don Williams] Interesting...I really liked Trigonometry and Geometry....
...specifically all this right triangle Trig stuff.[/QUOTE]

I know what you mean. I always thought the right triangle was the one you liked the best.

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It`s amazing that I struggled and studied to learn all that math just so I could forget it when I need it.

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PostPosted: Sun Mar 12, 2006 1:20 am 
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All that math to become a RAGBE- Right Angle Guitar Bridge Engineer!


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PostPosted: Sun Mar 12, 2006 2:31 am 
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[QUOTE=Joe Beaver] [QUOTE=Don Williams] Interesting...I really liked Trigonometry and Geometry....
...specifically all this right triangle Trig stuff.[/QUOTE]

I know what you mean. I always thought the right triangle was the one you liked the best. [/QUOTE]

I KNEW you thought that....

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PostPosted: Sun Mar 12, 2006 2:32 am 
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oh nevermind...double posted.Don Williams38788.4406481481

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PostPosted: Sun Mar 12, 2006 7:31 am 
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[QUOTE=Mike Mahar]
sine = hypotenuse / opposite side
cosine = hypotenuse / adjacent side
[/QUOTE]

I'm almost positive that you've inverted sine and cosine. The correct
definitions (as I remember them) are:

sin ? = opposite / hypotenuse
cos ? = adjacent / hypotenuse

(you correctly stated tan ? = opposite / adjacent).

I had a math teacher that taught me the mnemonic
SOH-CAH-TOA


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PostPosted: Mon Mar 13, 2006 12:16 am 
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Oops. You're right, they're backwards. Oddly enough, I'd never get that wrong when I do the math but writing it down in words somehow caused me to get it backwards.

[QUOTE=ecklesweb]    [QUOTE=Mike Mahar]
sine = hypotenuse / opposite side
cosine = hypotenuse / adjacent side
[/QUOTE]

I'm almost positive that you've inverted sine and cosine. The correct
definitions (as I remember them) are:

sin ? = opposite / hypotenuse
cos ? = adjacent / hypotenuse

(you correctly stated tan ? = opposite / adjacent).

I had a math teacher that taught me the mnemonic
SOH-CAH-TOA[/QUOTE]


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PostPosted: Tue Mar 14, 2006 12:13 am 
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Quit this math stuff. Blood is pouring from my ears and nose!


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