5.11. Given a triangle $A B C ; B D$ is a median, $\angle D B C=90^{\circ}$, $B D=\frac{\sqrt{3}}{4} A B$. Find $\angle A B D$.
5.11. According to the condition, $\angle D B C=90^{\circ}, B D=\frac{\sqrt{3}}{4} A B, A D=D C$ (Fig. 5.6); we need to find $\angle A B D=$ $=\alpha$. To find the angle $\alpha$, we will use the formula $\cos \alpha=\frac{\overline{B A} \cdot \overline{B D}}{|\overrightarrow{B A}||\overline{B D}|}$. Since $B D$ is the median of $\triangle A B C$,

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-602.jpg?height=214&width=573&top_left_y=785&top_left_x=563)

Fig. 5.6 then $B D=\frac{1}{2}(\overline{B A}+\overline{B C})$, from which $\overline{B A}=2 \overline{B D}-\overline{B C}$. Therefore,

$$
\cos \alpha=\frac{(2 \overline{B D}-\overline{B C}) \overline{B D}}{|\overline{B A}||| \overline{B D} \mid}=\frac{2 B D^{2}-\overline{B C} \cdot \overline{B D}}{|\overline{B A}||| B D}
$$

But $\overline{B C} \perp \overline{B D}$, i.e., $\overline{B C} \cdot \overline{B D}=0$, and from the equality $B D=\frac{\sqrt{3}}{4} A B$ it follows that $A B=\frac{4}{\sqrt{3}} B D$. Then the equality (1) will take the form

$$
\cos \alpha=\frac{2 B D^{2}}{\frac{4}{\sqrt{3}} B D^{2}}=\frac{\sqrt{3}}{2} \text {, from which } \alpha=30^{\circ}
$$

Answer: $30^{\circ}$.