Geometry/math Q

Heck, this post could fit several boards but since it pertains to a Modification I'll ask it here.

This should be real simple for the engineers and other math gifted people here. My geometry class was so long ago I barely remember this is possible.

What I have is a triangle (imaginary) with two sides that equal 4" and one side that equals 7.75". I need to know the degree of the angle opposite the long side.

Thanks.

Bones

It's been a real long time for me too. Seems like you need to find the length of the hypotenuse. That will divide your triangle into two right triangles, and you find the angles of those. Since you only asked for an equation, I'll let you solve for x.

Let me know if this works, it's been a long time and it's also 1:20 am.

HTH, Dan

working with that small of dimensions you could always draw it out then messure with a protractor.
That is the easiest way.

equalateral triangle....

angle a and b are both 14.36

angle c opposite the 7.75 leg is 151.27


why doesn't that sound right???

I think I screwed it up......

could be spobi.

Right on the money Beezil, 151. I think it's an Isosceles though. You still get a gold star.

oh, duh!!!!!!!

equalateral? why the hell did I say that! good thing eagle didn't read this post.....

Equalateral , Isosceles It's all the same at your age isn't it B?

It's that "oldtimers" disease....

You get used to it .

Thanks for the info guys!

Bones

........../|\
........./.|.\
......../..|..\....4" (c)
......./...|...\
....../....|....\
...../.....|b...\
..../......|......\
...--------------
...........|........|
.............3.875" (a)

a2 + b2 = c2 or 15.015625 +b2 = 16

b2=0.984375 b=0.9921567

COS(one half your angle) = 0.9921567 / 4

I don't have a cosine function on my calculator. You will have to take it from there. The angle is going to be pretty close to 170 degrees.
Hope this helps.
mattk

170 degrees?

mattk said:

........../|\
........./.|.\
......../..|..\....4" (c)
......./...|...\
....../....|....\
...../.....|b...\
..../......|......\
...--------------
...........|........|
.............3.875" (a)

a2 + b2 = c2 or 15.015625 +b2 = 16

b2=0.984375 b=0.9921567

COS(one half your angle) = 0.9921567 / 4

I don't have a cosine function on my calculator. You will have to take it from there. The angle is going to be pretty close to 170 degrees.
Hope this helps.
mattk

Click to expand...

Your method was correct, but the answer is not.

Beezil's 151.27 degrees was right.

The arccosine of 0.992/4 = 75.638
Times 2 = 151.27

It would have been easier to just use the sine function and take the known opposite over the hypontenuse.
sin(angle/2) = O/H
sin(angle/2) = 3.875/4 = 0.96875
Then you inverse the function and take the arcsin
arcsin(0.96875) = 75.638 = angle/2
angle = 151.27

C-ROC,

Just reading that made my head hurt. I knew there was a math god on here somewhere :worship: .

Bones :dunce:

Boy do I feel dumb, I'm sure that wheeling ability is'nt related to ones' education.

150 is pretty close to 170
Yes, wheeling ability is directly related to ones education. :lecture:
Bones, does this actually have anything to do with wheeling? What are you building?

151 degrees is a common a-frame angle amongst spanking machine builders.

bones, what are you using for the paddle wheel?

Bones said:

C-ROC,

Just reading that made my head hurt. I knew there was a math god on here somewhere :worship: .

Bones :dunce:

Click to expand...

haha, math god...pshhhhhh...that's basic trig. Talk to my blind math teacher, he's a math god. Imagine doing all this stuff without a calculator, or even a pencil and paper, it's quite amazing.

151.27 just sounds right, and is right from the measurement you gave. You have 2 sides which almost equal the length of the other side, meaning you have 2 tiny angles at the base of the triangle, and one large one at the peak. quick trig lesson for future reference and I'm bored as hell right now.

Only applies to right triangles (one 90* angle)

SOH (Sin of the angle [noted as sin 30* or whatever angle it is] = Opposite side / Hypotenuse side[opposite from 90* angle])
CHA (Cos = Hyp side / Adjacent side)
TOA (Tan = Opp side / Adj. side)

Just say the 3 - 3 capital letter things up there, it's a little way to remember it all. kinda like so-ka-toa.

If you have 2 sides, and no angles, you have to use the inverse of the sin, cos, tan (otherwise known at arcsin, arccos, arctan) function to find the angle jsut like they did to find the angle you needed...it's all on your calculator. If you understand this stuff you should find any length or angle you need. hehe, have fun. _nicko_

Cos is Adajacent over Hypotenuse (A/H) not H/A.

Sine = O/H; SOH
Cosine = A/H; CAH
Tangent = O/A; TOA

SOH-CAH-TOA

So the "ka" from your "so-ka-toa" should be a "cah"