Find dimension of triangle

[Almeidammiguel]

Hello i have one question i dont even know how to start the problem.
i attached the written problem!

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.

 

[Dr.Peterson]

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?

 

[Almeidammiguel]

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.

Thats why i discrimined the tan [MATH]\theta=\frac{a}{b}[/MATH] because the grid cannot change!

 

[HallsofIvy]

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"!

 

[Dr.Peterson]

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.