The diagonals of a rhombus, given area and tangent The area of the rhombus $ABCD$ is $24$ $cm^2$, if $\tan\measuredangle ABC=\dfrac{24}{7}$, find the diagonals $AC$ and $BD$.
I think we can say that $\measuredangle ABC$ is an acute angle. Is that true? Then $$\begin{cases}\tan\beta=\dfrac{24}{7}\\\sin^2\beta+\cos^2\beta=1\end{cases}$$ gives $\cos\beta=\dfrac{7}{25},\sin\beta=\dfrac{24}{25}.$ The area of $ABCD$ is $$S_{ABCD}=a^2\sin\beta=24\\a^2\cdot\dfrac{24}{25}=24\\a=5>0.$$ Now the Cosine Rule in triangle $ABC$ gives $$AC^2=2\cdot5^2-2\cdot5^2\dfrac{7}{25}=36,AC=6$$ The relationship $d_1d_2=48$ (from the area with the formula $S_{ABCD}=\frac{d_1d_2}{2}$) is very "clear". Can we come up with something else with the diagonals to make the solution better?

Let $\angle ABC=\theta$. Then $\angle ABD=\frac{\theta}{2}$. Suppose $AC\cap BD=P$ and $AC=2y$, $BD=2x$. In triangle $\triangle ABP$, we have $\tan\frac{\theta}{2}=\frac{y}{x}$. We also know that $(2x)(2y)=48\implies xy=12$. Using the tangent condition:
$$\frac{24}{7}=\frac{2(y/x)}{1-(y/x)^2}$$
Letting $y/x=z$, we have that $z=\frac{3}{4}$, hence $y=\frac{3}{4}x$, implying the result.
