A right-angled triangle has one leg of $465 \mathrm{~m}$, and the acute angle adjacent to it is $51.6^{\circ}$. Provide approximate values rounded to meters for the other two sides without using a function table, by applying the approximate formula learned in problem 1151. - Calculate the sides precisely as well.[^0]


[^0]:    ${ }^{1}$ K. M. L. $25(1962 / 12) 214$. p.
In the right-angled triangle $ABC$, let $AC = b = 465 \, \text{m}$, $\angle BAC = \alpha = 51.6^\circ$, $BC = a$, $AB = c$, and $\angle ABC = \beta = 38.4^\circ$. We denote the radian measure of the angles by $\alpha$ and $\beta$, which does not cause any confusion.

![](https://cdn.mathpix.com/cropped/2024_05_02_d9615feaebf26de00626g-1.jpg?height=328&width=376&top_left_y=264&top_left_x=861)

I. Using the mentioned approximate formula for both acute angles:

$$
\frac{3a}{2c + b} = \alpha, \quad \frac{3b}{2c + a} = \beta
$$

we obtain a system of linear equations for $a$ and $c$:

$$
3a - 2\alpha c = \alpha b, \quad \beta a + 2\beta c = 3b
$$

From this, we get:

$$
a = \frac{\alpha}{\beta} \cdot \frac{\beta + 3}{\alpha + 3} \cdot b, \quad c = \frac{9 - \alpha \beta}{2\beta(\alpha + 3)} b,
$$

The values of $\alpha$ and $\beta$ can be obtained with an error of $0.04\%$ using the approximation $\pi \approx 22/7$. On the other hand, the first significant digit of $a$ and $c$ is expected to be in the hundreds place, so we need to calculate to three significant figures. Therefore, in the initial steps of the preparatory calculations, we calculate to five, then only four significant figures:

$$
\begin{gathered}
\alpha \approx \frac{11 \cdot 51.6}{630}, \quad \beta \approx \frac{11 \cdot 38.4}{630} \\
a = \frac{51.6}{38.4} \cdot \frac{11 \cdot 38.4 + 1890}{11 \cdot 51.6 + 1890} \cdot 465 \approx \frac{119320}{94372} \cdot 465 \approx 1.264 \cdot 465 \approx 588 \, \text{m}, \\
c = \frac{9 \cdot 630^2 - 11^2 \cdot 51.6 \cdot 38.4}{2 \cdot 11 \cdot 38.4 (11 \cdot 51.6 + 1890)} \cdot 465 \approx \\
\approx \frac{3332300}{2076200} \cdot 465 \approx 1.605 \cdot 465 \approx 746 \, \text{m}
\end{gathered}
$$

II. Alternatively, using the function table, $a = b \tan \alpha \approx 465 \cdot 1.262 \approx 586.8 \, \text{m}$,

$$
c = \frac{b}{\cos \alpha} \approx \frac{465}{0.6211} \approx 1.610 \cdot 465 \approx 748.6 \, \text{m}
$$

According to these calculations, the error in neither of our calculations exceeds $0.5\%$.

Imre Bárány (Budapest-Mátyásföld, Corvin Mátyás g. I. o. t.)

Remarks. 1. It is sufficient to apply the approximate formula to the smaller angle - because this results in a smaller error - and we can replace the other equation with the Pythagorean theorem. This, however, leads to a quadratic equation, making the calculation more complex:

$$
\begin{array}{cl}
\beta a + 2 \beta c = 3b, & c^2 - a^2 = b^2 \\
a = \frac{b}{3\beta} \left(-3 \pm 2 \sqrt{9 - 3\beta^2}\right), & c = \frac{b}{3\beta} \left(6 \mp \sqrt{9 - 3\beta^2}\right)
\end{array}
$$

and we need to select the correct solution from the two. From this, we get $a \approx 586$ and $c \approx 747$, which indeed deviate even less from the exact values.

György Ferenczi (Budapest, I. István g. II. o. t.)

2. Many solutions were made difficult to verify due to the early substitution of numerical data. - Some cleverly used the fact that with good approximation $\alpha \approx 0.900$ and $\beta \approx 0.670$.
3. See also the 2nd remark attached to problem 1229.[^0]

[^0]:    ${ }^{1}$ In this issue, p. 133.