7. The external angles of a triangle are in the ratio $9: 16: 20$. From the vertex of the largest internal angle, the angle bisector and the altitude to the opposite side are drawn. What is the measure of the angle between the bisector and the altitude of the given triangle?

MINISTRY OF SCIENCE, EDUCATION AND SPORT OF THE REPUBLIC OF CROATIA AGENCY FOR EDUCATION AND UPBRINGING

CROATIAN MATHEMATICAL SOCIETY

## SCHOOL/CITY COMPETITION IN MATHEMATICS
7. According to the condition of the problem, $\alpha_{1}: \gamma_{1}: \beta_{1}=9: 16: 20$, we conclude that $\alpha_{1}=9 \mathrm{k}, \quad \gamma_{1}=16 \mathrm{k}$, $\beta_{1}=20 \mathrm{k}$.

The sum of the measures of the exterior angles of a triangle is $\alpha_{1}+\gamma_{1}+\beta_{1}=360^{\circ}$, or $9 \mathrm{k}+16 \mathrm{k}+20 \mathrm{k}=360^{\circ}$.

Solving this equation, we find that $\mathrm{k}=8^{0}$.

2 POINTS

It follows that $\alpha_{1}=9 \cdot 8^{0}=72^{0}, \gamma_{1}=16 \cdot 8^{0}=128^{0}, \beta_{1}=20 \cdot 8^{0}=160^{\circ}$.

1 POINT

The measures of the interior angles of the given triangle are respectively:

$\alpha=180^{\circ}-72^{0}=108^{0}, \quad \beta=180^{\circ}-160^{\circ}=20^{\circ}, \quad \gamma=180^{\circ}-128^{\circ}=52^{0}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_6e21dae8fd7ad64a6f32g-08.jpg?height=608&width=1128&top_left_y=1529&top_left_x=247)

In the right triangle $\triangle A E C$, we have $\gamma+|\Varangle E A C|=90^{\circ}$, so $|\Varangle E A C|=38^{\circ}$. 2 POINTS

Finally, $|\Varangle D A E|=\frac{\alpha}{2}-|\Varangle E A C|=54^{\circ}-38^{\circ}=16^{\circ}$.

$1 \mathrm{POINT}$

## SCHOOL/CITY COMPETITION IN MATHEMATICS  January 17, 2013.  8th grade - solutions

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