The area of the rhombus $ABCD$ is 2. A circle is inscribed in the triangle $ABD$, touching the side $AB$ at point $K$. A line $KL$ parallel to the diagonal $AC$ of the rhombus is drawn through point $K$ (point $L$ lies on the side $BC$). Find the angle $BAD$, given that the area of the triangle $KLB$ is $a$.

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Triangle $K B L$ is similar to triangle $A B C$. The similarity coefficient is $\frac{\sqrt{S_{K B L}}}{\sqrt{S_{A B C}}}=\sqrt{a}$.

Therefore, if $Q$ is the point of intersection of the diagonals of the rhombus, then

$\sin \angle B A C=\sin \angle B A Q=\frac{B Q}{A B}=\frac{B K}{A B}=\sqrt{a}$.

Thus, $\angle B A C=\arcsin \sqrt{a}, \angle B A D=2 \arcsin \sqrt{a}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_14311972f49418f31424g-16.jpg?height=365&width=715&top_left_y=1967&top_left_x=680)

## Answer

$2 \arcsin \sqrt{a}$.