$\left.\begin{array}{cc}{\left[\begin{array}{l}\text { Law of Cosines } \\ {[\text { Similar triangles (other) }}\end{array}\right]}\end{array}\right]$

In triangle $ABC$, $AC=2\sqrt{3}$, $AB=\sqrt{7}$, $BC=1$. Outside the triangle, a point $K$ is taken such that segment $KC$ intersects segment $AB$ at a point different from $B$, and the triangle with vertices $K, A$, and $C$ is similar to the original triangle. Find the angle $AKC$, given that angle $KAC$ is obtuse.
Prove that $\angle A K C=\angle C$.

## Solution

$A C$ is the largest side of triangle $A B C$. Since angle $K A C$ is obtuse, then $\angle K A C=\angle B$. Angle $C$ is not equal to angle $K C A$, as it is larger. Therefore,

$\angle A K C=\angle C$.

By the cosine theorem, $\cos \angle C=\frac{A C^{2}+B C^{2}-A B^{2}}{2 A C \cdot B C}=\frac{\sqrt{3}}{2}$.

Therefore, $\angle A K C=\angle C=30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96d74c2697cd905bd948g-22.jpg?height=309&width=623&top_left_y=2175&top_left_x=723)

## Answer

$30^{\circ}$.

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