12.406 The tangent of the angle between the median and the altitude drawn to the lateral side of an isosceles triangle is $\frac{1}{2}$. Find the sine of the angle at the vertex.
## Solution.

$\boldsymbol{B}$

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-290.jpg?height=190&width=322&top_left_y=1509&top_left_x=130)

A
Let $A M$ and $A D$ be the median and altitude of the given $\triangle A B C, A B=B C$. Then $\sin \angle B=\frac{A D}{A B}$, $A B^{2}=A D^{2}+B D^{2}, B D=|B M \pm D M|$ (the signs depend on the position of point $D$ relative to points $B$ and $M$ ),
$D M=A D \cdot \operatorname{tg} \angle D A M=\frac{A D}{2}, B M=\frac{A B}{2} \Rightarrow A B^{2}=A D^{2}+\left|\frac{A B}{2} \pm \frac{A D}{2}\right|^{2} \Rightarrow$ $\Rightarrow 5\left(\frac{A D}{A B}\right)^{2} \pm 2\left(\frac{A D}{A B}\right)-3=0 \Leftrightarrow 5 \sin ^{2} \angle B \pm 2 \sin \angle B-3=0 \Rightarrow\left[\begin{array}{l}\sin \angle B=1, \\ \sin \angle B=\frac{3}{5}\end{array}\right.$.

Answer: 1 or $\frac{3}{5}$.