10) Given a triangle with sides 5 m, 5 m, and 2.5 m. Find the angle α.

Solution:

This is an isosceles triangle with two sides of length 5 m. The angle \( \alpha \) is opposite to the base of length 2.5 m. Let's drop an altitude from the vertex between the two equal sides to the base. This altitude bisects the base and the vertex angle.

We have two right-angled triangles with hypotenuse 5 m and one leg \( 2.5 / 2 = 1.25 \) m.

Let's use the sine rule to find the angle \( \alpha \) in the original triangle:

\( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \)

Here, let the sides be \( a = 2.5 \text{ m} \), \( b = 5 \text{ m} \), \( c = 5 \text{ m} \). The angle \( \alpha \) is opposite to side \( a \).

We also have an angle \( \beta \) opposite to the sides of length 5 m.

Using the Law of Cosines to find \( \alpha \):

\( a^2 = b^2 + c^2 - 2bc \cos(\alpha) \)

\( (2.5)^2 = 5^2 + 5^2 - 2 \cdot 5 \cdot 5 \cos(\alpha) \)

\( 6.25 = 25 + 25 - 50 \cos(\alpha) \)

\( 6.25 = 50 - 50 \cos(\alpha) \)

\( 50 \cos(\alpha) = 50 - 6.25 \)

\( 50 \cos(\alpha) = 43.75 \)

\( \cos(\alpha) = \frac{43.75}{50} = 0.875 \)

\( \alpha = \arccos(0.875) \approx 28.96^{\circ} \)

Answer: \(\alpha \approx 28.96^{\circ}\)