My Math Forum Solving for sides and angles of a triangle.

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 July 14th, 2011, 12:36 AM #1 Newbie   Joined: Jul 2011 Posts: 16 Thanks: 0 Solving for sides and angles of a triangle. Hi. I'm working on a webpage that will 'solve' a triangle for a computer science summer course. The idea is that any information you give the page can be used to find the other pieces of the triangle. Unfortunately, I was never good at trig. What I'm wondering is: what information can I find depending on what information I have, which formulas I want to use in each case, and whether AAA can be used to find a ratio. As I understand it, if I know information X I can find information Y using method Z. Please correct me in all cases where I am wrong. Info X -[finds]- Info Y -[using]- Method Z 1 side (a) - Nothing - N/A. 1 side (a) and its' opposite angle (A) - Nothing? - N/A 1 side (a) and an angle touching it (B||C) - Nothing? - N/A 1 side (a) and any 2 angles (A+B||B+C||A+C) - Everything? - 180° rule. Sine law. Sine law. 2 sides (a + b) - Nothing? - N/A 2 sides (a + b) and the triangle's area - Everything? - *Heron's formula. Cosine Law. Sine Law. 180° rule.* 2 sides (a + b) and the angle trapped between them (C) - Everything? - Law of Cosine. Law of Sine. 180° rule. 2 sides (a + b) and an angle touching only one of them (A||C) ->If known angle < 90° - Everything?? --->WHATDO??? (This confuses the hell out of me). ->If known angle = 90° - Everything? - Pythagorean Theory. Sine Law. 180° rule. ->If known angle > 90° - Everything? - Sine Law. Sine Law. 180° rule. 3 sides (a + b + c) - Everything? - Cosine Law. Sine Law. 180° rule. 1 angle (A) - Nothing - N/A. 2 angles (A + B) - Third angle. **Ratio of side lengths?** - 180° rule. ??? 3 angles (A + B + C) - **Ratio of side lengths?** - 180° rule. ??? *I'm not sure if this is feasible/possible. I think it involves solving for a variable to the power of 4… but this should be possible on the calculator javascript is right?* **Although it's impossible to figure out the lengths of the sides from this it should be possible to determine their ratio and eventually express it as relative to a variable right? Ex: A=1.12x B=x C=1.6x (x>0). Any ideas how I would go about doing this?**
 July 14th, 2011, 12:51 AM #2 Newbie   Joined: Jul 2011 Posts: 16 Thanks: 0 Re: Solving for sides and angles of a triangle. I should clarify what I meant in the Heron's Formula note... Heron's formula is a formula for finding the area of a triangle using the lengths of its' sides. SQRT[S(S-A)(S-B)(S-C)] = area of triangle Where A = Length of side A of a triangle, B = Length of side B of a triangle, C = Length of side C of a triangle, and S = 'Demi-perimeter' or (A+B+C)/2. As an example of what I had in mind, I'll use a well known triangle: Assume we know side lengths A and B of a triangle, and the area of that triangle. We should be able to solve for C... A = 4 B = 3 Area = 6 SQRT[S(S-4)(S-3)(S-C)] = 6 Square both sides. [S(S-4)(S-3)(S-C)] = 36 Two variables won't do... Use the original equation for S to solve for C in terms of S. S=(A+B+C)/2 2S = 9+C 2S-9=C Replace C with it. [S(S-4)(S-3)(S-(2S-9))] = 36 [S(S-4)(S-3)(-S+9)] = 36 I don't really want to tackle the rest by hand because it's unlikely I could solve something to the power of 4 (but JS can solve for variables so it might work). Sorry if I made a bunch of mistakes there, I'm not much of a mathematician.
 July 15th, 2011, 12:08 PM #3 Newbie   Joined: Jul 2011 Posts: 16 Thanks: 0 Re: Solving for sides and angles of a triangle. No replies? I put in a little research on the triangle solving front, and I feel pretty confident with the list I made. However, I couldn't find anything on solving sides of length in an AAA problem based on a common variable with values restricted to greater than 0. Nor was I successful in tracking down any examples of reverse Heron's Formula in solving for a side of a triangle. Any help with either of these two issues would be greatly appreciated. As it were, I had made a mistake in my above write up. Since A = 4 and B = 3 S=(A+B+C)/2 2S = 7+C 2S-7=C [S(S-4)(S-3)(S-(2S-7))] = 36 [S(S-4)(S-3)(-S+7)] = 36 Which when run through the lovely tool here gives: 84S + -61S^2 + 14S^3 + -1S^4 = 36 Though it cannot provide a solution... (any ideas?) Anyhow substituting the value we know to be true for S in (6), does indeed yield the given area of the triangle. 84(6) + -61(6)^2 + 14(6)^3 + -1(6)^4 = 36 And the correct length for side C. 2S = A+B+C 12 = 7+C C = 7 If whatever I find to simplify the expression yields multiple S values then any negative values, or zero values can be eliminated. So I need 1. A way to calculate the ratio of the sides of a triangle given only their angles (if this data is insufficient then please tell me that). 2. Something that can solve for values of a variable based on an equation to the fourth degree.
 July 15th, 2011, 07:44 PM #4 Newbie   Joined: Jul 2011 Posts: 16 Thanks: 0 Re: Solving for sides and angles of a triangle. Okay so I think I know how to tackle my first question now, it's not too hard, even for an idiot like me. Since I want the lengths to be given in terms of a common variable, it's probably best for the smallest side(s) to always just be that variable (I'l use 'x' for my purposes). And of course we know that the smallest side of a triangle is always opposite the smallest angle… So I have my page checking for the smallest angle given and assigning the side opposite it the value of (1)x. From there we can use the Sine Law to find the other values in terms of x. Example: The angles of a triangle are as follows: A = 30° | B = 60° | C = 90° Give the lengths of it's sides as a value of x where x>0. (Note: While Pythagorean theory could be used in this instance, there's no point in programming both methods, so I'll be solving using Sine Law for each side since it will work for all triangles). The smallest angle is A so we know it's opposite side (Side a) is the smallest. We assign side a the value 1 (this is for js' benefit, it will be replaced with x in the end). Sine law states that: $\frac{a}{sin A}= \frac{b}{sin B} = \frac{c}{sin C}$ Since we've decided a = 1: $\frac{1}{sin 30}= 2$ $2= \frac{b}{sin 60} = \frac{c}{sin 90}$ $b= 2sin(60) = 1.73205...\ or\ sqrt{3}$ $c= 2sin(90) = 2$ $a= x\ |\ b = sqrt{3}\ |\ c = 2x$ $Given\ x > 0$ Which is of course the golden hemieq (30/60/90) triangle side length ratio $(1:sqrt{3}:2)$ I'm still completely stumped on how to solve for all possible values of a variable using an equation with that variable to the fourth degree... I'd really appreciate any suggestions on that front.
 July 16th, 2011, 01:35 AM #5 Newbie   Joined: Jul 2011 Posts: 16 Thanks: 0 Re: Solving for sides and angles of a triangle. Woops, 2 posts ago C clearly was equal to 5, not 7. 1 post ago $b=sqrt{3}x$ rather than just $b= sqrt{3}$. As for solving quartics: I found a site that handles it pretty well. When fed $x(x-4)(x-3)(-x+7)= 36$ it gives me $x=6\ or\ x=1$ I'm kind of amazed x=1 works, but I suppose it can be true if C is negative... anyhow I can screen the solutions by eliminating those where $((a+b)/2)\ >\ x$ as in all those cases c must be negative and the triangle fails to exist... I think. I still can't really picture a negative length side churning out a positive area, but that doesn't really matter. At this point the only problem will be using javascript to go through that form and take back the answer, but even that isn't too daunting. I'll leave this open for a few days in case someone sees any other potential problems with deciding between multiple possible values of x (there really should only be one value that makes sense though). I didn't really get a response to this, but solving it myself is better in a way I suppose. If nothing else, I appreciate being introduce to LaTeX by you guys; it's awesome! Cheers, Andrew

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