Author Topic: Trignometry  (Read 13181 times)

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alanjt

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Re: Trignometry
« Reply #15 on: August 10, 2011, 02:07:35 pm »
Oh my GAWD.just answer the question already.

The answer is YES.

There problem solved....now we can get on with the rest of the excitement in our lives!

 :-D :-D
But Sorry; certifiable proof is needed along with your first born that you are "really good" at trigonometry.
Wait, we actually accomplished something here?
Not yet by my simple standards.  But getting closer.
2+2=5
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BlackBox

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Re: Trignometry
« Reply #16 on: August 10, 2011, 02:14:17 pm »
Better idea, why doesn't the OP just get 'good'... 'really good' at trigonometry on their own:

Learn trigonometry at Khan Academy
"Potential has a shelf life." - Margaret Atwood

alanjt

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Re: Trignometry
« Reply #17 on: August 10, 2011, 02:15:07 pm »
Better idea, why doesn't the OP just get 'good'... 'really good' at trigonometry on their own:

Learn trigonometry at Khan Academy
Khaaaaaan!
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kdub

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Re: Trignometry
« Reply #18 on: August 10, 2011, 10:55:14 pm »
 :-)
Anybody good at trig here, I mean really good?

yes.


edit: sorry, I should have read the complete thread before posting ... seems the question has  been answered.
called Kerry in my other life

Sometimes the question is more important than the answer.

alanjt

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Re: Trignometry
« Reply #19 on: August 10, 2011, 10:57:26 pm »
Anybody good at trig here, I mean really good?

yes.
HaHa. Sentence structuring and grammar.
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Keith™

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Re: Trignometry
« Reply #20 on: August 10, 2011, 11:10:27 pm »
You guys are seriously afflicted ...

Now that the question has been answered in the affirmative .. if you haven't seriously pissed off yet another newt swamper, then perhaps at some point we can get him to develop his thoughts more.

I guess everyone here already knows everything and they never had the occasion to have to ask anyone for anything. It must be a nice bubble you are living in.
Proud provider of opinion and arrogance since November 22, 2003 at 09:35:31 am
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Jeff H

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Re: Trignometry
« Reply #21 on: August 11, 2011, 12:08:26 am »
I am guilty and apologize,
 
Hopefully he understands that it had nothing to do with him and will reply.
He has not been active since all these posts and would be great if they disappeared.

alanjt

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Re: Trignometry
« Reply #22 on: August 11, 2011, 12:10:04 am »
I am guilty and apologize,
 
Hopefully he understands that it had nothing to do with him and will reply.
He has not been active since all these posts and would be great if they disappeared.
Oh come on, post #5 is solid gold.  :roll:
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mjfarrell

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Re: Trignometry
« Reply #23 on: August 11, 2011, 12:35:38 am »

I guess everyone here already knows everything and they never had the occasion to have to ask anyone for anything.
Yes, I've had to ask, and I do so by asking a question that is derived to achieve the answer I needed, not find out if someone might have the ability or aptitude to answer.  Because there is a difference.

Note:

Is anyone here good at trig, I mean really good

VS

Can someone tell me how to solve for or provide a solution to calculating the included angle in the following trigonometric equation? (Enter your equation here)

Which form do you believe will get me the answer or the process to answer the mathematical challenge I am being pressed with?
Be your Best


Michael Farrell
http://primeservicesglobal.com/

BlackBox

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Re: Trignometry
« Reply #24 on: August 11, 2011, 01:10:47 am »
I am guilty and apologize,
 
Hopefully he understands that it had nothing to do with him and will reply.
He has not been active since all these posts and would be great if they disappeared.
Oh come on, post #5 is solid gold.  :roll:

Khan trumps Star Trek... Especially given Keith's moaning.  :wink:
"Potential has a shelf life." - Margaret Atwood

caddman6425

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Re: Trignometry
« Reply #25 on: August 11, 2011, 02:07:01 am »
Thanks all, I guess I really deserved this.  The question that I need help with is really long and drawn out, so I was thinking that before I tried to type everything out, I'd better find out if there was some people available right at the present to help.  I thought that I'd notified when I got answers, but I guess not, and I got so caught up trying to solve the problem that I forgot to check. Sorry all.  I'm attaching a drawing and I guess well start from there.  I'm trying to sole "o"

Jeff H

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Re: Trignometry
« Reply #26 on: August 11, 2011, 02:46:55 am »
Are trying to find the arch length between 2 points is so using trim
 
 

caddman6425

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Re: Trignometry
« Reply #27 on: August 11, 2011, 03:20:28 am »
This is why I say that it is going to take some time to explain what I'm trying to accomplish.  There is four sections to the drawing. The first is the text, the second is the portion of the drawing that looks like a mess.  The two parell arcs represent a arched steal beam.   I'm trying to come up with an equation so as to solve the plum cut of the beam at any point along 1/2 the periphery of the arc.  The triangle that is formed by the beam depth and the unknown vertical cut, forms an oblique triangle, the oppsite side is not tangent to the arc.  The only knowns are the beam depth which is 10 and the adjcent angle of 19.47 deg.  I can sovle the right triangle with the angle an Hyp as well as the opp side, but I can sole the other angle of the oblique triangle since I don't have enough knowns.  This is where I need help, I've stared so close at all the trees, that I don't see the forest anymore.  I know that it is solvable, but I can't figure where the relatinships are to get me where I want to go.  There is a lot more, but I hate typing so I guess I'll go with the flow as it comes along.

SEANT

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Re: Trignometry
« Reply #28 on: August 11, 2011, 03:38:52 am »
I think that this is what you want:

o = SQRT((TAN(A)*10)^2 – (SIN(A)*10)^2)

Which should be 1.1785

The drawing has it at 1.1419, but the angle between the radius “c” and “h” is not 90 degrees.  Should it be 90?
Sean Tessier
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caddman6425

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Re: Trignometry
« Reply #29 on: August 11, 2011, 03:48:17 am »
No it should not be 90, it isn't tangent to the arc.  The vertical line that intersect the top arc is th plum cut length.  When you form the triangle that small side is the intersection points of the triangle.  If it was tangent, then I'd have no problem.  You're equation is base on the ang being tangent or at 90 deg