In triangle $ABC$, the sides are $AB=18$, $BC=12$, and $CA=8$ units, while in triangle $ACD$, the other two sides are $CD=7$ and $DA=6$ units. The line $AC$ separates points $B$ and $D$. According to Péter, the angle $BAD$ is acute, while Pál claims it is obtuse. - Can the dispute be resolved? (No tools other than the school function table may be used.)
Using the cosine theorem for triangles $ABC$ and $ACD$, and rounding to four decimal places:

$$
\begin{aligned}
& \cos \angle BAC = \cos \alpha = 61 / 72 = 0.8472, \\
& \cos \angle CAD = \cos \delta = 17 / 32 = 0.5312
\end{aligned}
$$

From which the interpolated part of the angles, expressed as common fractions, in degrees:

$$
\alpha + \delta = 32.0 + \frac{8}{90} + 57.9 + \frac{2}{150} = 89.9 + \frac{46}{450} = 90^{\circ} + 8''
$$

This supports Pál's statement.

![](https://cdn.mathpix.com/cropped/2024_05_02_62caf495898ef759c96eg-1.jpg?height=309&width=653&top_left_y=599&top_left_x=714)

However, using the logarithmic cosine table:

$$
\begin{gathered}
\log \cos \alpha = 9.9280 - 10 \\
\log \cos \delta = 9.7253 - 10 \\
\alpha + \delta = 32.0 + \frac{4}{50} + 57.9 + \frac{1}{120} = 89.9 + \frac{53}{600} = 90^{\circ} - 42''
\end{gathered}
$$

According to this, Péter would be correct.

The contradiction arises because the data in our table are rounded, and the errors introduced into the calculation are larger for the small absolute difference between $90^{\circ}$ and $\alpha + \delta$.

We can resolve the debate, for example, by calculating the value of $\cos (\alpha + \delta)$ using the addition theorem. If this value is positive, then Péter's statement is correct; if negative, then Pál's; and if zero, then the angle in question is a right angle. For this:

$$
\begin{gathered}
\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{72^2 - 61^2} / 72, \quad \sin \delta = \sqrt{32^2 - 17^2} / 32 \\
72 \cdot 32 \cos (\alpha + \delta) = 61 \cdot 17 - \sqrt{1463 \cdot 735}
\end{gathered}
$$

The square of the first term is $3721 \cdot 289 = 1075369$, and the square of the term to be subtracted is $1463 \cdot 735 = 1075305$, which is smaller. Therefore, $\cos (\alpha + \delta) > 0$, and the angle in question is an acute angle.

Judit Bod (Budapest, Apáczai Csere J. Practical Grammar School III. Class)

Remarks. 1. We can decide without using the addition theorem based on the size relationship between the sum $s = \cos^2 \alpha + \cos^2 \delta$ and 1. If $s \gtreqless 1$, then $\cos^2 \delta \gtreqless \sin^2 \alpha = \cos^2 (90^{\circ} - \alpha)$, and $\delta \lesseqgtr 90^{\circ} - \alpha$, because for positive acute angles, a larger cosine value corresponds to a smaller angle. In our case, $s = (244^2 + 153^2) / 288^2 = 82945 / 82944 > 1$, so $\alpha + \delta$ is an acute angle.

2. According to a multi-digit table, $90^{\circ} - \alpha - \delta < 3''$.