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To find the smallest angle of the triangle given sides 8 cm, 5 cm, and an angle of 90° between them, we will use the cosine rule and basic trigonometric principles.
First, denote the sides of the triangle as follows:
- \( a = 8 \) cm (side opposite to angle A)
- \( b = 5 \) cm (side opposite to angle B)
- \( C = 90^\circ \) (angle between sides \( a \) and \( b \))
To find angle C (the right angle), the smallest angle in this case:
\[ \angle C = 90^\circ \]
To find angle A (opposite side \( a \)):
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
Calculating \( c \) using the Pythagorean theorem:
\[ c = \sqrt{a^2 + b^2} = \sqrt{8^2 + 5^2} = \sqrt{64 + 25} = \sqrt{89} \]
Now, substitute the values into the cosine rule:
\[ \cos A = \frac{5^2 + (\sqrt{89})^2 - 8^2}{2 \cdot 5 \cdot \sqrt{89}} \]
\[ \cos A = \frac{25 + 89 - 64}{2 \cdot 5 \cdot \sqrt{89}} \]
\[ \cos A = \frac{50}{10 \cdot \sqrt{89}} \]
\[ \cos A = \frac{5}{\sqrt{89}} \]
Calculate \( \cos A \):
\[ \cos A \approx 0.532 \]
Now, find angle A:
\[ \angle A = \cos^{-1}(0.532) \]
\[ \angle A \approx 58.5^\circ \]
Finally, to determine the smallest angle of the triangle:
\[ \text{Smallest angle} = \angle C = 90^\circ \]
Therefore, the smallest angle of the triangle is \( \boxed{90^\circ} \).