In a triangle, the length of side $AB$ is $10 \mathrm{~cm}$, the length of side $AC$ is $5.1 \mathrm{~cm}$, and $\angle CAB = 58^{\circ}$. Determine $\angle BCA$ to the nearest hundredth of a degree.
Solution. In a triangle, using the usual notation, we write the cosine rule for side $a$:

$$
a^{2}=b^{2}+c^{2}-2 b c \cdot \cos \alpha
$$

By repeatedly applying the cosine rule, we can write:

$$
c^{2}=a^{2}+b^{2}-2 a b \cdot \cos \gamma
$$

From this,

$$
\cos \gamma=\frac{a^{2}+b^{2}-c^{2}}{2 a b}
$$

Substitute the expression for $a^{2}$ from (1) into (2):

$$
\cos \gamma=\frac{b^{2}+c^{2}+b^{2}-2 b c \cdot \cos \alpha-c^{2}}{2 b \sqrt{b^{2}+c^{2}-2 b c \cdot \cos \alpha}}
$$

Combine and simplify the fraction by $2 b \neq 0$. We get:

$$
\cos \gamma=\frac{b-c \cdot \cos \alpha}{\sqrt{b^{2}+c^{2}-2 b c \cdot \cos \alpha}}
$$

Then substitute the given values and perform the indicated operations: $\cos \gamma \approx -0.02346$. The cosine value is negative, so the angle we are looking for is greater than $90^{\circ}$. The value of $\gamma$ is $180^{\circ}-88.6557^{\circ} \approx 91.35^{\circ}$.

Juhász Anikó (Eger, Gárdonyi G. Gymnasium, 12th grade)

Note. Many incorrect and incomplete solutions were received for this problem. If we first calculate the length of side $a$ using the cosine rule:

$$
a^{2}=b^{2}+c^{2}-2 b c \cdot \cos \alpha=5.1^{2}+10^{2}-2 \cdot 5.1 \cdot 10 \cdot \cos 58^{\circ}
$$

then $a^{2} \approx 71.958$ and from this $a \approx 8.48 \, \text{cm}$. If we then write the sine rule, we get:

$$
\sin \gamma=\frac{\sin 58^{\circ}}{a} \cdot c
$$

The first surprise comes after substitution: if we substitute the approximate value of $a$, then $\sin \gamma \approx 1.000056$ is obtained, which is not possible. Before we think that the triangle does not exist, let's consider that the triangle can be constructed without any further issues. The problem lies elsewhere.

When we rounded $a$ to two decimal places, 8.48 is smaller than the actual value, thus reducing the denominator in the sine rule and increasing the value of the fraction and thus $\sin \gamma$. The consequence of rounding is usually the inaccuracy of the final result, but in this case, $\gamma$ is in a critical range: approximately $90^{\circ}$, with a sine value close to 1. The rounding, therefore, changes not the value of the result but its nature. It also becomes clear that $\sin \gamma$ is approximately 1, indicating that the triangle is nearly a half-cut equilateral triangle, with an angle of about $90^{\circ}$ at vertex $C$.

Let's calculate more precisely. With four decimal places, $a \approx 8.4828$ and then $\sin \gamma \approx 0.9997$.

If someone gets this far, they only need to determine whether $\gamma$ is an acute or obtuse angle. $\gamma$ is the largest angle in the triangle, but nothing more can be deduced from the given data. This can be clarified using the cosine rule.

![](https://cdn.mathpix.com/cropped/2024_05_02_d93feee26c4d303e08c4g-1.jpg?height=246&width=499&top_left_y=1991&top_left_x=802)