I have to solve the next problem:
Given H (relative height to the hypotenuse) and R (radius of the circle inscribed in the triangle) of a rectangle triangle, can you calculate the value of its interior angles from the H/R relation.
But I don't understand what does "relative height to the hypotenuse" means.
Problem source: http://coj.uci.cu/24h/problem.xhtml?abb=1335
First, relative height to the hypotenuse means the height that's perependicular to the hypotenuse and pass through the vertex of the right angle.
Now for the problem. Because the triangle is right we can apply the Pythagorean Theorem. So we have:
$$a^2 + b^2 = c^2$$ $$a^2 + 2ab + b^2 = c^2 + 2ab$$
We know that $$Hc = ab$$ so we make a substitution:
$$(a+b)^2 = c^2 + 2cH$$ $$a + b = \sqrt{c^2 + 2cH}$$
Note that we are interested only in the positive value of the square root, because $a$ and $b$ have positive length. Now subtitute back in your equation for $\frac{H}{R}$. So we have:
$$\frac HR = \frac{a+b+c}{c}$$ $$\frac HR = \frac{\sqrt{c^2 + 2cH} + c}{c}$$
After some algebaric transformation we'll end up with:
$$c = \frac{2R^2}{H-2R}$$
Because $H$ and $R$ are known values, we can easily get c.
Now let the base of the altitide $H$ divide the side $c$ in two segments $p$ and $q$.
We know that:
$$p+q = c \quad \quad \text{ and } \quad \quad pq = H^2$$
So the the length of $p$ and $q$ are actually the solutions of the following quadratic equation. This follows from Vieta's formula.
$$x^2 - cx + H^2 = 0$$
Now we can find the side of the right triangle after applying Pythagorean Theorem:
$$a = \sqrt{q^2 + H^2}$$ $$b = \sqrt{p^2 + H^2}$$
And finally let $\alpha$ be the angle opposite of side $a$ and $\beta$ the angle opposite of side b. Then we have:
$$\alpha = sin^{-1} \frac ac$$ $$\beta = sin^{-1} \frac bc$$