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.

A HINT

Yes, if $$\tan\gamma =\frac{3}{4}$$ then the double angle formula gives $$\tan2\gamma =\frac{24}{7}$$.

$$\gamma$$ is the angle the sides make with $$BD$$.

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?