Question 380·Hard·Math

You can scale the whole triangle without changing its angles. Take $BC=2$. Since $ABC$ is equilateral, all its sides are equal, so $AB=AC=2$ too.

Draw the altitude from $A$ to $BC$, and call its foot $M$. In an equilateral triangle, this line bisects both the base and the $60^\circ$ angle at $A$. So $BM=MC=1$ and $m\angle BAM=30^\circ$.

Right triangle $ABM$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle. Its short leg is $BM=1$, so the special-triangle ratio makes its long leg $AM=\sqrt{3}$.

Let $d_1$ stand for $BD$. Substitute $BC=2$ into the given fraction:

Type $d_1/2\sim2-\sqrt{3}$; the $\sim$ tells Desmos to find $d_1$. Under PARAMETERS, it shows $d_1\approx0.536$. Since that is less than $BM=1$, $D$ lies between $B$ and $M$. Use the ratio, not the sketch, to locate $D$.

Let $\theta=m\angle MAD$. Because $D$ is between $B$ and $M$, $MD=BM-BD=1-d_1$. In right triangle $AMD$, tangent is opposite over adjacent:

Ray $AD$ lies between $AB$ and $AM$, so $\angle DAB$ is what’s left of $\angle BAM$ after removing $\theta$:

Subtract from the half-angle, not the full $60^\circ$ angle at $A$.

Set Desmos to Degrees. Type $30-\arctan((1-d_1)/\sqrt{3})$; inverse tangent gives $\theta$ from its tangent ratio. Desmos shows $15$, so $m\angle DAB=15^\circ$. Choice A.