Can the product of the hypotenuse and base of a right angled triangle be expressed in terms of the angle between the base and height?

Question

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.

Answer 1

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

• $A=1=\frac{RH}2=\frac{LR\cos\theta}2\implies LR=\frac{2}{\cos \theta}$

Answer 2

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.