In triangle $A B C$, from vertex $C$ the bisectors of the internal and external angles are drawn. The first bisector forms an angle of $40^{\circ}$ with side $A B$. What angle does the second bisector form with the extension of side $A B$?
The angle between the bisectors of adjacent angles is $90^{\circ}$.

## Solution

Let the specified bisectors intersect the ray $A B$ at points $K$ and $M$ respectively. Then $\angle M C K = 1/2 \angle A C B + 1/2 (180^{\circ} - \angle A C B) = 90^{\circ}$. Therefore, $\angle K M C = 90^{\circ} - \angle M K C = 90^{\circ} - 40^{\circ} = 50^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_05_06_67482594d9a3488694ecg-24.jpg?height=418&width=648&top_left_y=540&top_left_x=705)

## Answer

$50^{\circ}$.

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