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Geometry problem triangle

  • Thread starter late347
  • Start date
301
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1. Homework Statement
line m2-to-m is 3km longer than line m1-to-m

what are lengths for ##M~~M_2## and ##M~~M_1##
geometria kuva kolmio m m1 m2.jpg

2. Homework Equations
pythagorean theorem hopefully can be used


3. The Attempt at a Solution

geometria kuva kolmio edistynyt.png

use the picture to your advantage in hopefully creating a valid system of equations. With which, X and Y can both be solved.

However the question only asked really what was (X-3) and what was (X). Y should hopefully only be used as an intermediary variable to that purpose(???)

system of equations is as follows:
$$y^2~+~ 5.6^2~= ~x^2$$
$$(18-y)^2 ~+ ~5.6^2~ = ~(x-3)^2$$



The top equation can be reduced into following form...
##x= \sqrt{y^2+5.6^2}##
the lower equation can be reduced into the form of...
##y^2~-36y~+346.36= x*(x-6)##

insert top equation into the lower equation. Hopefully this was correct because on pen-paper-style it took an entire page, really....
##35y^2 -630y +2724.89=0##

from calculator
y = 10.7736967
y= 7.226303295

We can test these Y's and hopefully to ascertain whether or not the system is true.

I think that with the top value of Y (y ~ 10.773 .......)

the originatl system of equations seems to work ok.

testing can be done with calculator for example with assigning variables with some values such as
y=10.7736967
from the top eqution we plug in the y value, and calculate that x should be equal to about 12.14218022


with that x value, you can plug it into the lower equation and it should hold true.

so it looks like length of x = 12.14218022 km
it looks like length of (x-3) = 9.142180224 km
 
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feedback or critique is welcome. Especially if I was wrong ... (obviously...)

also alternative tactic of solving is welcome.

maybe this is dumb question but how can a person know for a fact. WHICH side from a given triangle with a 90deg angle, is supposedly the hypotenuse of that particular triangle? Without knowing strictly the lengths of sides at first instance???
 

haruspex

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There is an easier way. If you take the difference of your two original equations the quadratics cancel, leaving you a linear relationship between x and y.
how can a person know for a fact. WHICH side from a given triangle with a 90deg angle, is supposedly the hypotenuse of that particular triangle? Without knowing strictly the lengths of sides at first instance?
I assume you mean, if you are given two side lengths and the fact that the triangle is right-angled, but don't know the relationship between that and the given lengths. If so, you cannot tell. In a2+b2=c2, you can plug in any two values for a and b and solve, or plug the larger in as c and the smaller in as a and solve.
 
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Well I googled it and it looks like in a right angle triangle...

Even if you dont know any lengths... then hypotenuse ought to be that length which is opposite to the right angle (the right angle's location you woyld know initially)

I did not understand what you meant by difference of squares though
 

haruspex

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it looks like in a right angle triangle..
What does?
hypotenuse ought to be that length which is opposite to the right angle
Of course.
I did not understand what you meant by difference of squares though
In your first equation, you have a y2 on the left and x2 on the right. If you expand the parentheses in the second equation that will also have those squared terms. If you subtract the first equation from the second (left from left, right from right) you will have an equation with no quadratic terms in x and y.
 
301
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What does?

Of course.

In your first equation, you have a y2 on the left and x2 on the right. If you expand the parentheses in the second equation that will also have those squared terms. If you subtract the first equation from the second (left from left, right from right) you will have an equation with no quadratic terms in x and y.
I see now, but yet I wonder whether or not that is really so much easier in actual fact.

I think my teacher used to say that theres about 4 main ways to solve systems of equations.

Placement method (gauss's method??). He never really had time to explain or demonstrate that method in class, because our progress in math class was somewhat at a slow rate. I think we only really had some homework problems with systems of equations. Not those problems in the last exam.

summation method (reduce one of the equations in the pair, into suitable form, and multiply by (-1) and sum the top equation into the bottom equation and the certain variables cancel out)

"equating to each other" method (reduce both equaations until they are the of same form such as x= [something].... Set the something part to equal the other something part) In the end it is a form of placement, though, I suppose.

"matrices" method. I once knew how to do it but I have long since forgotten how it is supposed to work.
 

haruspex

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I see now, but yet I wonder whether or not that is really so much easier
You did not post all your working, but it looked like you would get a quartic equation that would take some work to simplify. My method avoids that, never being worse than a quadratic.
theres about 4 main ways to solve systems of equations
Those methods, as general methods, are for linear equations; these are not.
The equating method does work here, and is what you used. I suppose you could consider my method as an example of the summation method.
 
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nvm.
 

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