# It's So Sad, , ,
**URL:** https://boards.straightdope.com/t/its-so-sad/4999
**Category:** Factual Questions
**Created:** [November 28, 1999, 10:01pm UTC](https://boards.straightdope.com/t/its-so-sad/4999 "1999-11-28T22:01:00Z")
**Posts on this page:** 18
**Page:** 1
### Author:  [@DougC](https://boards.straightdope.com/u/DougC)
#### Post date: [November 28, 1999, 10:01pm UTC](https://boards.straightdope.com/t/its-so-sad/4999/1 "1999-11-28T22:01:00Z")
, , , but I forgot how to do trig. (It’s been 12+ yrs since hi-skool)
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- I need to know how to find the angles of a right triangle when only the sides are known. I looked at about 20 sites and none _seemed_ to tell how; they only say how to find an opposite side or angle from two sides and one known angle (the angle that’s not 90 deg). - I hit buttons on a calculator for about 45 minutes and I can’t seem to arrive at anything close to the angles of a 3-4-5 triangle. Sites that claim to be able to tell how all seem to charge for the privilege. - MC
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### Author:  [@funneefarmer](https://boards.straightdope.com/u/funneefarmer)
#### Post date: [November 28, 1999, 10:33pm UTC](https://boards.straightdope.com/t/its-so-sad/4999/2 "1999-11-28T22:33:00Z")
(a_a)+(b_b)=(c\*c)
Where a and b are legs and c is the hypotenuse (sp)
At least I think I remember properly
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### Author:  [@funneefarmer](https://boards.straightdope.com/u/funneefarmer)
#### Post date: [November 28, 1999, 10:35pm UTC](https://boards.straightdope.com/t/its-so-sad/4999/3 "1999-11-28T22:35:00Z")
oops…
Angles !
Slamming my head against the monitor.
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### Author:  [@Ringo](https://boards.straightdope.com/u/Ringo)
#### Post date: [November 28, 1999, 10:38pm UTC](https://boards.straightdope.com/t/its-so-sad/4999/4 "1999-11-28T22:38:00Z")
Law of cosines:
a²=b²+c²-2bccosA
where a, b & c are the lengths of the sides and A is the angle opposite side a, B is the angle opposite side b and C is the angle opposite side c.
I got A=90º B=53.13º and C=36.87º
If you go to [the bighub](http://www.thebighub.com/) and search for law of cosines you’ll find lots of free instruction.
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### Author:  [@Monty](https://boards.straightdope.com/u/Monty)
#### Post date: [November 28, 1999, 10:39pm UTC](https://boards.straightdope.com/t/its-so-sad/4999/5 "1999-11-28T22:39:00Z")
MC:
In your OP above, the angles of the triangle aren’t unknown–one angle is known. Remember that one angle of a right triangle is 90 degrees (a right angle) and that the hypotenuse (the side opposite the right angle) must be the longest side of the triangle. With that information, you can use any of those sites to which you referred to get the other two angles.
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### Author:  [@funneefarmer](https://boards.straightdope.com/u/funneefarmer)
#### Post date: [November 28, 1999, 10:39pm UTC](https://boards.straightdope.com/t/its-so-sad/4999/6 "1999-11-28T22:39:00Z")
[http://www.xavierhs.org/departments/Mathematics/Trig/Rt%20Triangle/rttrig.htm](http://www.xavierhs.org/departments/Mathematics/Trig/Rt%20Triangle/rttrig.htm)
Try that.
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### Author:  [@Monty](https://boards.straightdope.com/u/Monty)
#### Post date: [November 28, 1999, 10:42pm UTC](https://boards.straightdope.com/t/its-so-sad/4999/7 "1999-11-28T22:42:00Z")
An alternative option is to go ahead and construct a triangle using the lengths of the sides given, assuming all three lengths are provided. Since triangles are inflexible, there’s only one solution to the construction and you can then measure the thing with a protractor.
But I’d go with what I described above.
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### Author:  [@BigRoryG](https://boards.straightdope.com/u/BigRoryG)
#### Post date: [November 28, 1999, 11:14pm UTC](https://boards.straightdope.com/t/its-so-sad/4999/8 "1999-11-28T23:14:00Z")
I’m not quite sure what you want here, buit it’s all ratios… SOH CAH TOA
Sine is the ratio of the opposite to the hypotenuse
Cosine is Adjacent over Hypotenuse and
Tanent is opposite over Adjacent
So, If you want the angle… what you need to enter into your calculator is {one of the trig buttons} {the predetermined ratio of the two sides that you have} and the answer you get will be the angle.
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“C’mon, it’s not even tomorrow yet…” - Rupert
If you need a graphic solution, [http:\ alk.to\Piglet](http://talk.to%5CPiglet)
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### Author:  [@system](https://boards.straightdope.com/u/system)
#### Post date: [November 29, 1999, 2:34am UTC](https://boards.straightdope.com/t/its-so-sad/4999/9 "1999-11-29T02:34:00Z")
I hope you all appreciate that I’m not even going to TRY to respond to the OP . . . .
-Melin
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### Author:  [@Ringo](https://boards.straightdope.com/u/Ringo)
#### Post date: [November 29, 1999, 2:41am UTC](https://boards.straightdope.com/t/its-so-sad/4999/10 "1999-11-29T02:41:00Z")
Of course, dear, I _know_ you finished that second lap in record time.
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### Author:  [@DougC](https://boards.straightdope.com/u/DougC)
#### Post date: [November 29, 1999, 3:44am UTC](https://boards.straightdope.com/t/its-so-sad/4999/11 "1999-11-29T03:44:00Z")
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- tanks peoples. I am trying to model a cube in 3-D, so protractors aren’t an option. Just for conversation’s sake, as I see it for each polar coordinate I only need angle X, angle Y and distance (distance for X and Y being the same). The one book I have on hand about 3-D modeling says to use angle and distance for all three. I do not understand; is that just to average errors? The cube will stay centered on the screen and rotate; so I am supposing I don’t need anything to find viewing angle. - MC
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### Author:  [@NanoByte](https://boards.straightdope.com/u/NanoByte)
#### Post date: [November 29, 1999, 3:44am UTC](https://boards.straightdope.com/t/its-so-sad/4999/12 "1999-11-29T03:44:00Z")
You mean engineers have sweat all this time to bring you people extremely low-cost calculators (well, at least when assembled in sweat shops), and you don’t know how to use them? This problem is a total no-brainer if you have a scientific calculator (hardware or software) – one that has arctangents on it.
If you know the lengths of the 3 sides of a right triangle and want to know the 2 angles besides the definition-determined 90-deg angle, all you have to do is take the ratio of the 2 shortest sides (i.e., divide them), either way around, and then take the arc- or inverse tangent of the result. Subtract this angle from 90 deg. The 90-deg angle is, of course, opposite the longest side of the triangle, the smallest of the angles found before and after the above subtraction is opposite the smallest angle, and the remaining angle is, then, obviously opposite the remaining side. Of course, trig tables will also get you there in the same manner. You don’t need any fancy formulae (unless you just like to calculate out numerous terms of a long series for an arctangent.
Ray (with plenty of angles for getting off on a tangent)
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### Author:  [@DougC](https://boards.straightdope.com/u/DougC)
#### Post date: [November 29, 1999, 4:06am UTC](https://boards.straightdope.com/t/its-so-sad/4999/13 "1999-11-29T04:06:00Z")
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- Actually I read it isn’t the engineers that sweat, it’s the Chinese schoolkids that assemble the calculators who do.
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- I was trying things that were too complicated; I thought there was more to it than there is. Usually I tend not to remember the formulas; I try remember a “working example” of a principle to use it. - MC
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### Author:  [@DougC](https://boards.straightdope.com/u/DougC)
#### Post date: [November 29, 1999, 4:09am UTC](https://boards.straightdope.com/t/its-so-sad/4999/14 "1999-11-29T04:09:00Z")
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- And I don’t have a trig table; I don’t bother with decorating much. As far as I know, this is just a regular computer desk. - MC
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### Author:  [@NanoByte](https://boards.straightdope.com/u/NanoByte)
#### Post date: [November 29, 1999, 7:25am UTC](https://boards.straightdope.com/t/its-so-sad/4999/15 "1999-11-29T07:25:00Z")
He who put computer on trig table show sines of soft wear.
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### Author:  [@glee](https://boards.straightdope.com/u/glee)
#### Post date: [November 29, 1999, 9:17am UTC](https://boards.straightdope.com/t/its-so-sad/4999/16 "1999-11-29T09:17:00Z")
‘All right, I’m in charge now. Who are you?’
‘Officer Pythagoras, sir.’
‘What do we know about the suspect?’
‘He shows sines of a recent tan, sir, but we don’t know if that angle is right.’
‘Pythagoras, gimme a bull horn!’
‘Yessir.’
‘ALL RIGHT NANOBYTE, WE KNOW YOU’RE IN THERE. THIS IS THE PUN POLICE. COME ON OUT WITH YOUR HANDS UP YER BUM.’
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### Author:  [@DougC](https://boards.straightdope.com/u/DougC)
#### Post date: [November 30, 1999, 6:56am UTC](https://boards.straightdope.com/t/its-so-sad/4999/17 "1999-11-30T06:56:00Z")
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- Does your version 4.5 Quickbasic have the help file QB45ADVR.HLP ? - Mine don’t; QB4.5 came with the computer on the system disk but that file isn’t there. I also note that the QB4.5 compiler downloadable from:
- [http://gameprog.ampleworks.com/qbasic/Features/compilers.html](http://gameprog.ampleworks.com/qbasic/Features/compilers.html)
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- doesn’t have it either. What gives? Was this one of Bill’s evil plans that the feds stopped in time? - MC
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### Author:  [@Harmonious\_Discord](https://boards.straightdope.com/u/Harmonious_Discord)
#### Post date: [November 30, 1999, 1:53am UTC](https://boards.straightdope.com/t/its-so-sad/4999/18 "1999-11-30T01:53:00Z")
The sine of the angle = oposite side divided by the hypotenuse.
sin(Angle)=op/hyp
or angle=arc sine(op/hyp)
The cosine of the angle = the adjacent side divided by the hypotenuse.
cos(angle)=adj/hyp
or Angle=arc cosine(adj/hyp)
The tangent of the angle = the oposite side divided by the adjacent.
tan(angle)=op/adj
or Angle=arc tangent(op/adj)
(a^2+b^2)^.5=c
^is to the power of
a & b are the sides forming the right angle
c is the hypotenuse