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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?

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3 Answers 3

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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.

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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 $BD$.

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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?

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