Find the total length of two line segments of two overlapping right triangles 
Find the length of $AE+EB$ ?
(A) $\frac{128}{7}$
(B) $\frac{112}{7}$
(C) $\frac{100}{7}$
(D) $\frac{96}{7}$
(E) $\frac{56}{7}$
My solution:
For $\Delta AEB$ :
$\angle BAC = sin^{-1}(\frac{5}{13}) = 22.62^{\circ}$
$\angle ABD = sin^{-1}(\frac{9}{15}) = 36.87^{\circ}$
$\angle AEB = 180 - \angle BAC - \angle ABD = 120.52^{\circ}$
Use Law of Sines:
$\frac{AE}{sin(\angle ABD)} = \frac{AB}{sin(\angle AEB)}$
$AE = \frac{AB}{sin(\angle AEB)} \times sin(\angle ABD)$
$AE = \frac{12 \times \frac{9}{15}}{sin 120.51} = 8,372$

$\frac{EB}{sin(\angle BAC)} = \frac{AB}{sin(\angle AEB)}$
$EB = \frac{AB}{sin(\angle AEB)} \times sin(\angle BAC)$
$EB = \frac{12 \times \frac{5}{13}}{sin 120.51} = 5.367$

$AE+EB = 8.372 + 5.367 = 13,7387 \approx \frac{96}{7}$ Answer: (D)
My question:
Is there a solution without having to compute both arcsin $\angle BAC$ & $\angle ABD$? The reason I'm asking is because the choices all have 7 as denominators. So I'm guessing there may be a solution that contains only integers.

Yes there is, drop a perpendicular down from $E$ to $AB$ and denote the point as $X$. Let $EX = x$ and $AX = 12-y$ and $BX = y$. Since $\triangle EBX $ ~ $\triangle DBA$, we have that $\frac9x = \frac{12}y$. Also, since we have $\triangle CBA$ ~ $\triangle EXA$, we have that $\frac5x = \frac{12}{12-y}$. Now, solving  for $x,y$ using our equations, we get $x=\frac{45}{14}, y = \frac{30}{7}$. Plugging in these values, we get $AX = \frac{54}{7}$ and $BX = \frac{30}{7}$ and $EX = \frac{45}{14}$. Now using the pythagorean theorem, we get $AE = \frac{117}{14}$ and $BE=\frac{75}{14}$. Now, adding these lengths, we get that the sum is $\frac{96}{7}$

