Quadratic equation and trigonometry

[Saitama]

You're right. I guess a triangle should indeed have angles strictly between 0 and 180 degrees.

In the edge case A is a straight angle. That is correct.
If we exclude the edge case, angle A is slightly less than a straight angle.

So the final answer is 3+4+5=12? :)

 

[I like Serena]

Yep!

 

[Saitama]

Yep!

Thanks a lot I Like Serena!

 

[Ray Vickson]

Thank you both for the participation.

Continuing with k=4+2cosB, I don't see how both the bounds are reachable. Clearly, k=2 when B=$\pi$ but since this is a triangle, we can't have this or else it is a straight line. For the upper bound i.e k=6, it is possible when B=0. Again this is not possible in a triangle. So the range is 2<k<6. Sorry if this sounds dumb.



I found cosine of A for the triangle you state, it doesn't look possible. cosA comes out to be -1.

http://www.wolframalpha.com/input/?...(sqrt(2)-1)^2-(sqrt(2)+1)^2)/(2*(sqrt(2)-1)*2)

The minimum of f is 2, and that corresponds to a triangle that has collapsed into a line segment, so is not really a triangle at all. However, if you take some (a,c) giving, say f = 2.00001, there will be a legitimate triangle giving that value of f. For f = 2 the (positive) integers <= f are 1 and 2. For f = 2.00001 the positive integers <= f are still just 1 and 2. That is why f_min = 2 gives a legitimate lower bound on the integer sum, even though no actual triangle achieves that value of f.

 

[skj91]

I believe everyone is overlooking the fact that the formula which provides the solutions to the quadratic equation in the form of: ax² + bx + c is actually not

b²-4ac

but in fact it is:x = (-b ± √(b² - 4ac)) / 2a The '±' in this particular formula is the most significant mathematical notation you'll ever meet as that is precisely how you can derive your upper and lower bounds. First substitute into that formula your values and record the result of subtracting the portion under the root from -b, then do the same thing again, only find the result of adding the value under the root sign to your -b.

That is assuming b represents your y-coordinate, a is your x-coordinate and c is your hypotenuse, which is the standard representation used.

 

[skj91]

Also when dealing with sin and cos in equations where you are solving for real values of x and y, you can and will have results which have a negative value (i.e. -1) due to the way sine waves are graphed and the concept of the 'unit circle' where 0° -- (which is at the same location as 360° which is a full circle of 2π length with a center point at the coordinate (0,0) and starting point of the arc swept by its radius at the coordinate (1,0) -- will yield a value of -1 for x when the arc formed by the radius pivots around the the (0,0) point by 180° and lands on π, horizontally aligned with the x-axis. Inversely you will get a value of -1 for y when it swings another 90° past that to the 270° mark vertically aligned with the y-axis. You will, however, never produce a negative value for the radius (which is also always representative of a triangle's hypotenuse, when graphed).

Or you could go with the difference or sum of sines rules but that's probably more of a headache to get through than it's worth . Go with the ± operation on the quadratic equation over that crazy stuff.

And don't forget Pythagoreas, you may need to make use of his:

a² + b² = c²

theorem at some point.

(or x² + y² = h² in this case)

 

[skj91]

mods, please delete this (accidental double posting)

 

[Ray Vickson]

I believe everyone is overlooking the fact that the formula which provides the solutions to the quadratic equation in the form of: ax² + bx + c is actually not

b²-4ac

but in fact it is:


x = (-b ± √(b² - 4ac)) / 2a


The '±' in this particular formula is the most significant mathematical notation you'll ever meet as that is precisely how you can derive your upper and lower bounds. First substitute into that formula your values and record the result of subtracting the portion under the root from -b, then do the same thing again, only find the result of adding the value under the root sign to your -b.

That is assuming b represents your y-coordinate, a is your x-coordinate and c is your hypotenuse, which is the standard representation used.

Sorry: nobody overlooked anything. The question asked about a triangle whose sides a, b and c are coefficients of a quadratic equation having a root of multiplicity 2. Aside from that, the actual roots of the equation have nothing else to do with the problem.

 

[skj91]

Sorry: nobody overlooked anything. The question asked about a triangle whose sides a, b and c are coefficients of a quadratic equation having a root of multiplicity 2. Aside from that, the actual roots of the equation have nothing else to do with the problem.

Right, well... I'm not about to go into explaining how the coefficients of which you speak are directly proportionate to the slope of the line formed from (0,0) to the identified (a,b) or (x,y) point.-- That formed line, of course being = to the radius on a unit circle which = to the hypotenuse of a graphed triangle which is = to rise over run = y/x = opposite/adjacent = sin θ /cos θ = tan θ and that the Quadratic formula method of solving for x and then applying that to solve for their y pairs does in fact apply the correct ratio of y to x values since those coefficients are what identify what that ratio is.

Anyway, back to the OP, is there some reason you can't approach this from another methodology, for example through the Law of of Sines? Or maybe you need to look at it from the perspective of the unit circle where you have actual values to plug into those sin leg or angle variables. It all really depends on what kind of answer you (or your boss/instructor/whoever) is expecting. Is an answer in the form of another equivalent variable for sin, cos, or tan, enough? Or are you expected to find the actual number values of a specific size of triangle? If the latter is the case then you would have had to have been given some kind of known values of the specified triangle, either leg lengths, angles in degrees or the pi locations of one angle's sin and/or cos or tan functiions or any of the exact values of sin, cos, or tan when mapped to a unit circle... some kind of other info must be given or calculating any actual solution simply can't be done.


p.s. The formula I previously referenced has a misplaced parenthesis and should actually read as:

-b ± √(b² - (4ac)/2a)

 

[Saitama]

skj91, please have a look at the problem statement again. Your posts has nothing to do with solving the given problem.

 

[skj91]

It has everything to do with it, you simply aren't aware of trigonometric nor quadratic functions. It's called a SINE wave for a reason, you know. A graph of a triangle is in reality just a straight line dropped from endpoint of any radii of any circle with an origin of (0,0) wherever it happens to be resting in any of the 4 quadrants of a co-ordinate system. If you're comprehension of mathematical concepts is so limited that can't even recognize that there are multitudes of methods which can be used to come to to the same exact results then you're beyond my ability to help, but that doesn't mean you should just go and assume that I'm talking nonsense. It's not my fault if you don't know what I'm talking about. Again something that would have been no problem if you had provided a bit more info about your inquiry.

I can only assume then that this is really a simple geometric problem in which case you cross-multiply and all your sins cancel out leaving you with an answer of 1/1 which is simply 1 but I don't think that's what your actually being asked to do which is why I'm asking... don't you have any known values to work with?

Good luck.

 

[Saitama]

It has everything to do with it, you simply aren't aware of trigonometric nor quadratic functions. It's called a SINE wave for a reason, you know. A graph of a triangle is in reality just a straight line dropped from endpoint of any radii of any circle with an origin of (0,0) wherever it happens to be resting in any of the 4 quadrants of a co-ordinate system. If you're comprehension of mathematical concepts is so limited that can't even recognize that there are multitudes of methods which can be used to come to to the same exact results then you're beyond my ability to help, but that doesn't mean you should just go and assume that I'm talking nonsense. It's not my fault if you don't know what I'm talking about. Again something that would have been no problem if you had provided a bit more info about your inquiry.

I can only assume then that this is really a simple geometric problem in which case you cross-multiply and all your sins cancel out leaving you with an answer of 1/1 which is simply 1 but I don't think that's what your actually being asked to do which is why I'm asking... don't you have any known values to work with?

Good luck.

Did you have a look at the problem?

I am completely aware of most of the stuff, so please, think or at least have a look at the previous posts by me and the other posters.

 

[skj91]

Yes I looked at the problem and still you have not identified what you're trying to solve for or what method you are being asked to solve it with. You mention finding the upper and lower limits but that is impossible to do without having actual numbers to work with. In other words WHAT ARE YOUR GIVEN VALUES FOR ANY OF THE 6 SIDES OR ANGLES? It's far too difficult to attempt to explain solving equations with nothing but variables the whole way through so don't expect anyone to be willing to try doing that.
Both the equations you offer for starting points are not complete equations. SinA/SinC + SinC/SinA is not an equation, it's a meaningless pair of terms. What is it = to? What are you solving for? It can't be an equation unless there is an '=' sign somewhere so you're not giving us all the information you have to work with, here. You mention that it is a quadratic equation but then you reject using the actual "Quadratic Formula" method of solving it. That leaves only three other algebraic level methods as options.



1) Solve through Completing the square. To use this you must put your equation into a factored form. (Since you won't divulge your known values I can't explain how this is done).


2) Solve through Square Root method. Your Initial equation would have to be in the form of:

(ax-b)²-c²=0​

to use this method.


3) Solve by Factoring.

Sin²B = (sinA/sinC + SinC/SinA)² →
(SinA/SinC +SinC/SinA)(SinA/SinC +SinC/SinA) →
Sin²A/Sin²C + 2(SinASinC/SinCSinA) + Sin²C/Sin²A = Sin²B​

The only other way is to use the Quadratic formula which I have already pointed out that you're using only half of the formula required to use that method.

You tagged this thread as Trigonometry and Quadratic so I obviously assumed your question was at a college trigonometry level. Obviously that is not the case or my last 2 replies would have made perfect sense to you. This is basic high-school algebra... tag it as such.