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I've been working on a math problem recently whose small subpart part is this. I don't want to post the whole problem and be spoon fed it, but I've been struggling with this sub part of it and since my math skills are still trivial the solution may require maths which I have to learn so,

Can the product $\mathtt(LR)$ where L is the hypotenuse of a right angled triangle and R is it's base be expressed using trigonometric relations of only $\theta$? Where $\theta$ is the angle between the hypotenuse and height H of the right angled triangle?

If yes derive the expression? Otherwise prove it not possible.

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  • $\begingroup$ Well...the trig functions of $\theta$ don't change if you scale the triangle but your product... $\endgroup$ – lulu Mar 24 '18 at 14:45
  • $\begingroup$ How could it be since that product is different for the similar triangles $3-4-5$ and $9-12-15$.? The ratio is invariant. Maybe you should tell us a little about the problem that led you to this unsolvable one. $\endgroup$ – Ethan Bolker Mar 24 '18 at 14:45
  • $\begingroup$ it's not possible to use only an angle, there are infinitely many right triangles that have the same angles $\endgroup$ – Vasya Mar 24 '18 at 14:47
  • $\begingroup$ @EthanBolker it has to do with integrating the surface area of a cone $\endgroup$ – noobAtMaths Mar 24 '18 at 14:57
  • $\begingroup$ @noobAtMaths Let ask an OP with the original problem on the cone $\endgroup$ – user Mar 24 '18 at 14:59
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Note that indicating with L the hypotenuse and with R the base we have

$$R=L\sin \theta \implies LR=L^2 \sin \theta$$

Thus it depends from $L^2$, and in general from a length$^2$. It is equal to $\sin \theta$ only if $L=1$.

Note that it is possible only if we fix some kind of constraint as for example the area of the triangle $=1$ that is, indicating with $H$ the height

  • $A=1=\frac{RH}2=\frac{LR\cos\theta}2\implies LR=\frac{2}{\cos \theta}$
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  • $\begingroup$ It still contains L? it should only contain trigonometric functions of $\theta$ $\endgroup$ – noobAtMaths Mar 24 '18 at 14:46
  • $\begingroup$ @noobAtMaths I've shown the relation precisely to explain that it is not possible since $LR$ is a length$^2$ in general we need some other length to define it. For example I can use $R^2+S^2$ instead of $L^2$ but I Always need some length to express the quantity. In other words it is not adimensional. $\endgroup$ – user Mar 24 '18 at 14:50
  • $\begingroup$ Yes I see it wasn't obvious for me prior to your edit. $\endgroup$ – noobAtMaths Mar 24 '18 at 14:51
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Assuming that by "base" you mean the side forming $\theta$ with the hypotenuse.

We have

$$\cos\theta =\frac{L}{R}$$

(or $\sin$ instead if you meant the other side). Thus $$LR= R^2\cos\theta.$$ Now, assume there is an expression $f(\theta)$ depending only of $\theta$ that equals $LR$. We have $$f(\theta)=LR=R^2\cos\theta,$$ thus $f(\theta)$ depends on $R$, which is a contradiction. Therefore such an expression doesn't exists.

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