Given diagonals in a parallelogram. Find sides. Given a parallelogram with $d_1 = AC = 26$ cm, $d_2 = BD = 18$ cm and $\sin \displaystyle \angle AOD = \frac{12}{13}$.
Find $AB = a$ and $AD = b$.

What I did in order to solve it

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*Using the formula for the area

$S = \frac{d_1d_2\sin \displaystyle \Phi}{2} = \frac{26 \times 18 \times 12/13}{2} = 216$

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*Using Heron's formula for $\triangle BOC$ to find $b$
Defining $OC = a = 26 / 2 = 13$, $OB = b = 18 / 2 = 9$, $BC = c$
$p = \frac{a + b +c}{2} = \frac{13 + 9 + c}{2} = \frac{22 + c}{2}$
$S = \sqrt{p(p-a)(p-b)(p-c)}$
I believe when I find $b$, I will able to find $a$ too, but kinda stuck at that point.

$\cos AOD =\pm \sqrt{1-\sin^2 AOD}=\pm \sqrt{1-(12/13)^2}=\pm 5/13$.
Since $AOD < 90^\circ$, $\cos AOD = 5/13$
So $b^2=AO^2+OD^2-2 AO\cdot OD \cos AOD = 13^2+9^2-2\cdot 13\cdot 9\cdot (5/13)=160$.
Since $\sin DOC=\sin(180^\circ-AOD)=\sin AOD=12/13$, we also have $\cos DOC=\pm 5/13$, but this time we choose the minus sign since $DOC>90^\circ$, so $\cos DOC = -5/13$
So $a^2=DO^2+OC^2-2 DO\cdot OC \cos DOC = 9^2+13^2-2\cdot 9\cdot 13\cdot (-5/13)=340$.
