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?

If $O$ is the intersection of $AC,BD$
$2AO=AC,2BO=BD$
$24=\dfrac{AC\cdot BD}2$
$2\angle ABO=\angle ABC=2x$(say)
Now $\tan x=\dfrac{AO}{BO}$
But $\tan2x=\dfrac{24}7>0\implies0<2x<\dfrac\pi2$
Use $\tan2x=\dfrac{2\tan x}{1-\tan^2x}$ to find out $\tan x$
Can you take it home from here?
