B1. The vertices of triangle $ABC$ lie on a circle with a radius of $5 \, \text{cm}$ and divide it such that the ratio of the sizes of the interior angles of the triangle is $\alpha: \beta: \gamma = 2: 3: 4$.

a) Accurately calculate the length of the longest side to two decimal places.

b) The central sector whose chord is the longest side of the triangle is rolled into a cone. Accurately calculate the surface area of the cone's lateral surface.
B1.

a) In the ratio $\alpha: \beta: \gamma=2: 3: 4$, we introduce a new variable, say $t$, and consider that the sum of the interior angles of a triangle is $180^{\circ}$. We get the equation $2 t+3 t+4 t=180^{\circ}$, whose solution is $t=20^{\circ}$. Thus, we calculate the size of the angles and get the result $\alpha=40^{\circ}, \beta=60^{\circ}$, and $\gamma=80^{\circ}$. The longest side of the triangle is $c$, as it lies opposite the largest angle $\gamma=80^{\circ}$. Side $c$ is also a side of triangle $A B S$. The angle opposite side $c$ in triangle $A B S$ is the central angle of the inscribed angle $\gamma=80^{\circ}$ and measures $\angle A S B=160^{\circ}$. Triangle $A B S$ is an isosceles triangle with legs of $5 \mathrm{~cm}$ and a vertex angle of $160^{\circ}$. Using the cosine rule $c^{2}=5^{2}+5^{2}-2 \cdot 5 \cdot 5 \cdot \cos 160^{\circ}$ or the trigonometric function $\sin \frac{160^{\circ}}{2}=\frac{\frac{c}{2}}{5}$, we calculate the side $c \doteq 9.85 \mathrm{~cm}$.

b) The lateral surface area of the cone is equal to the area of the circular sector. The circular sector has a central angle of $160^{\circ}$ and a radius of $5 \mathrm{~cm}$. The area of the circular sector is calculated using the formula $S_{i}=\frac{\pi r^{2} \alpha_{s}}{360^{\circ}}=\frac{\pi 5^{2} \cdot 160^{\circ}}{360^{\circ}}$ and we get $S_{p l}=\frac{100 \pi}{9} \mathrm{~cm}^{2}$.

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The realization that the area of the circular sector is equal to the surface area of the cone's lateral surface ...................... 1 point

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