## 7. Obtuse triangle

Given is an obtuse triangle $\triangle A B C$ where the measures of all angles are expressed in degrees as natural numbers. The angle bisector of the obtuse angle $\angle B A C$ intersects the side $\overline{B C}$ at point $D$, and the foot of the altitude from vertex $A$ is point $N$. Point $D$ is located between points $N$ and $B$. The measure of angle $\angle C B A$ is ten times the measure of angle $\angle D A N$, which is also a natural number when expressed in degrees. What is the largest possible measure of the obtuse angle of this triangle in degrees?

Result: $\quad 158$
## Solution:

Let $\alpha=|\Varangle \mathrm{BAC}|, \quad x=|\Varangle D A N|$.

Then $|\Varangle \mathrm{A} B \mathrm{C}|=10 x$.

Since triangle $\triangle A B N$ is a right triangle, we have $\frac{\alpha}{2}+x+10 x=90^{\circ}$, or $\alpha+22 x=180^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_43f5242198ec4dde97b4g-5.jpg?height=526&width=696&top_left_y=2030&top_left_x=1231)

Since the smallest possible $x$ is $1^{\circ}$, the largest possible $\alpha$ is $158^{\circ}$.