Bakayev E.v.

Inside an isosceles triangle $\$ \mathrm{ABC} \$$, a point $\$ K \$$ is marked such that $\$ C K=\mathrm{AB}=\mathrm{BC} \$$ and $\angle K A C=30^{\circ}$. Find the angle $\$ A K B \$$.

#
![](https://cdn.mathpix.com/cropped/2024_05_06_5ff42d6d57f9dccbb1f5g-16.jpg?height=51&width=1927&top_left_y=2473&top_left_x=40)
line $BC$

![](https://cdn.mathpix.com/cropped/2024_05_06_5ff42d6d57f9dccbb1f5g-17.jpg?height=389&width=634&top_left_y=0&top_left_x=715)

![](https://cdn.mathpix.com/cropped/2024_05_06_5ff42d6d57f9dccbb1f5g-17.jpg?height=51&width=1955&top_left_y=406&top_left_x=40)
of triangle $\$ \mathrm{ABC}, \mathrm{AC}>\mathrm{BC}$. Therefore, $\angle A B C>60^{\circ}$, which means $\$ \mathrm{~L} \$$ and $\$ \mathrm{~B} \$$ lie on opposite sides of \$ \$L lies on the smaller arc \$AC\$. Thus, the inscribed angle \$CAL \$ is half of the central angle \$CBL\$, which is $30^{\circ}$.

Obviously, the point $\$ K \$$ that satisfies the conditions of the problem is unique, so it coincides with the point symmetric to $\$ \mathrm{~L} \$$ with respect to side $\$ \mathrm{AC} \$. Therefore, triangle $\$ \mathrm{AKL} \$$ is equilateral, and point $\$ K \$$, like point $\$ B \$$, lies on the perpendicular bisector of segment $\$ \mathrm{AL} \$. Hence, $\langle \mathrm{AKB}=\langle LKB=180^\{\circ\} - \frac{1}{2} \backslash$ angle $A K L=150 \wedge\{$ circ $\}$.

## Answer

$150^{\circ}$.