Find dimension of triangle

Almeidammiguel

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May 21, 2020
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Hello i have one question i dont even know how to start the problem.
i attached the written problem!
6fc60e51-fc46-44ce-acca-8234be691fdb.jpg
Its given an angle [MATH]\theta [/MATH]and a dimension s.
i want to know x being #1 and #2 two integer numbers such that it makes a right triangle with sides #1*s, #2*s and x.
 
Let's see if I can make sense of what you've said; the grammar of the last sentence is not at all clear.

You show a right triangle with legs [MATH]4s[/MATH] and [MATH]5s[/MATH] in the picture; I suppose the numbers 4 and 5 could be any integers, which you are calling #1 and #2. Let's call them [MATH]a[/MATH] and [MATH]b[/MATH] instead, so that in the picture [MATH]a = 4[/MATH] and [MATH]b = 5[/MATH]; and I'll assume that they, too, are given, in addition to length [MATH]s[/MATH] and angle [MATH]\theta[/MATH].

The expression you wrote becomes [MATH]\frac{as}{\sin(\theta)} - \frac{bs}{\cos(\theta)}[/MATH]. Can you explain your thinking? Is this supposed to equal x?

One problem I see is that [MATH]a[/MATH] and [MATH]b[/MATH] will determine [MATH]\theta: \tan(\theta) = \frac{bs}{as} = \frac{b}{a}[/MATH]. So if [MATH]\theta[/MATH] is given, then you are assuming that its tangent is a rational number, which is not true in general. If you knew [MATH]a[/MATH] and [MATH]b[/MATH], rather than [MATH]\theta[/MATH], you could find [MATH]\theta[/MATH] from them.

Can you explain the requirements more fully?
 
Let's see if I can make sense of what you've said; the grammar of the last sentence is not at all clear.

You show a right triangle with legs [MATH]4s[/MATH] and [MATH]5s[/MATH] in the picture; I suppose the numbers 4 and 5 could be any integers, which you are calling #1 and #2. Let's call them [MATH]a[/MATH] and [MATH]b[/MATH] instead, so that in the picture [MATH]a = 4[/MATH] and [MATH]b = 5[/MATH]; and I'll assume that they, too, are given, in addition to length [MATH]s[/MATH] and angle [MATH]\theta[/MATH].

The expression you wrote becomes [MATH]\frac{as}{\sin(\theta)} - \frac{bs}{\cos(\theta)}[/MATH]. Can you explain your thinking? Is this supposed to equal x?

One problem I see is that [MATH]a[/MATH] and [MATH]b[/MATH] will determine [MATH]\theta: \tan(\theta) = \frac{bs}{as} = \frac{b}{a}[/MATH]. So if [MATH]\theta[/MATH] is given, then you are assuming that its tangent is a rational number, which is not true in general. If you knew [MATH]a[/MATH] and [MATH]b[/MATH], rather than [MATH]\theta[/MATH], you could find [MATH]\theta[/MATH] from them.

Can you explain the requirements more fully?

The expression i wrote i was just thinking in how to start the problem! Dont mind it it might not be anything usefull!

The requirement is: if i have a grid of circles and rotate this grid being the rotation center one circle, somewhere there will be a circle that is completly aligned in the axis direction with the circle that serves as rotation centre - Image in attachment. 38d1d2ce-24fa-4ed7-a720-1ff7bdd950de.jpg

Thats why i discrimined the tan [MATH]\theta=\frac{a}{b}[/MATH] because the grid cannot change!
 
It looks to me like this is over determined. You have legs of length 4s and 5s and angle \(\displaystyle \theta\) opposite the 5s side. From that \(\displaystyle tan(\theta)= \frac{5s}{4s}= \frac{5}{4}\) so that \(\displaystyle \theta= 51.3\) degrees yet you say that \(\displaystyle \theta\) is "given"!
 
It appears that you are missing the fact that your s is irrelevant, since the angle depends only on the ratio so that s cancels out. There is going to be "somewhere there will be a circle that is completely aligned" if the tangent of your angle happens to be a rational number. That is true only for very special angles. Going out further to look doesn't help, because of the cancellation.
 
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