7.2. The external angles of a triangle are proportional to the numbers $5: 7: 8$. Find the angle between the altitudes of this triangle drawn from the vertices of its smaller angles.
Answer: $90^{0}$.

Solution. The sum of the exterior angles of a triangle is

$$
180^{\circ} \cdot 3-180^{\circ}=360^{\circ}
$$

From this, we get that one of the exterior angles of the triangle is $\frac{360^{\circ}}{5+7+8} \cdot 5=90^{0}$, and therefore the adjacent interior angle of the triangle is $90^{0}$. Consequently, the other two angles are the smaller angles of the triangle, and the heights drawn from them coincide with the legs of the triangle, the angle between which is $90^{\circ}$.

Comment. An answer without justification - 0 points.