A triangle has one side measured as 441 meters, and the angles on it are measured as $16.2^{\circ}$ and $40.6^{\circ}$. Provide approximate values rounded to meters for the other two sides, without using a function table, by using problem 1 of problem 1151. - For comparison, also calculate the sides precisely.[^0]


[^0]:    ${ }^{1} 1962 / 158$. page.
In triangle $ABC$, let $AB = c = 441 \, \text{m}$, $\angle BAC = \alpha = 16.2^\circ$, $\angle ABC = \beta = 40.6^\circ$, and $BC = a$, $CA = b$. The foot of the perpendicular from $B$ is $B_1$, $BB_1 = m$, and $AB_1 = c_1$, so $\angle ABB_1 = \beta_1 = 73.8^\circ$.

I. Apply the formula examined in problem 1151 to the right triangle $ABB_1$ for both angles (denote the radian measure of the angles with the corresponding Greek letter; this does not cause confusion):

$$
\frac{3m}{2c + c_1} = \alpha, \quad \frac{3c_1}{2c + m} = \beta_1
$$

This gives us a system of linear equations for the legs $m$ and $c_1$:

$$
3m - \alpha c_1 = 2c \alpha, \quad -\beta_1 m + 3c_1 = 2c \beta_1
$$

From which we get:

$$
m = \frac{2c \alpha (\beta_1 + 3)}{9 - \alpha \beta_1}, \quad c_1 = \frac{2c \beta_1 (\alpha + 3)}{9 - \alpha \beta_1}
$$

Here,

$$
\alpha \approx \frac{16.2 \cdot 11}{630} \approx 0.2828, \quad \beta_1 \approx \frac{73.8 \cdot 11}{630} \approx 1.288
$$

In the subsequent calculations, we will also write 4 significant digits: $m \approx 123.9$, $c_1 \approx 431.9$.

![](https://cdn.mathpix.com/cropped/2024_05_02_269f04dd67dcfea2b631g-1.jpg?height=402&width=992&top_left_y=932&top_left_x=556)

Now we calculate the hypotenuse $a$ and the leg $CB_1 = a_1$ of the right triangle $BCB_1$ from the leg $m$ and the angles $\angle CBB_1 = \beta_2 = \beta_1 - \beta = 73.8^\circ - 40.6^\circ = 33.2^\circ$ and $\angle BCB_1 = \gamma_1 = \alpha + \beta = 56.8^\circ$. This procedure is similar to that in exercise 825, with appropriate changes in notation:

$$
a_1 = \frac{\beta_2}{\gamma_1} \cdot \frac{\gamma_1 + 3}{\beta_2 + 3} \cdot m, \quad a = \frac{9 - \beta_2 \gamma_1}{2 \gamma_1 (\beta_2 + 3)} \cdot m
$$

From this, $a_1 \approx 87.6$, $a \approx 147.0$, so $b = c_1 - a_1 \approx 344$.

II. By precise calculation using the sine rule, $a = 147 \, \text{m}$ and $b = 343 \, \text{m}$ are obtained.

Gyula Kulcsár (Pannonhalma, Benedekine Order III. Class)

Remarks. 1. The solutions arrived at the result through various methods. Many used, for example, the Pythagorean theorem. Some wrote direct formulas for $a$ and $b$, but it was much more common to substitute the numerical data early, which reduces the clarity of the calculations. Note, for example, that after calculating the product $\alpha \beta_1$, it is easy to calculate $\alpha (\beta_1 + 3)$ and $\beta_1 (\alpha + 3)$, and this repetition occurs with $\beta_2 \gamma_1$, $\beta_2 (\gamma_1 + 3)$, and $\gamma_1 (\beta_2 + 3)$.

2. Our problem asked for the determination of the sides "using the 1151st problem." In principle, a more accurate approximate value would be obtained if, in addition to the approximate formula appearing there, the table of the difference between the exact and approximate values of the angles (given in approximately $10^\circ$ steps) were also taken into account. For example, it would have been useful to reduce the radian measure of an angle around $40^\circ$ by $8 \cdot 10^{-4}$. [^0]

[^0]:    ${ }^{1}$ See p. 145.