Given a triangle $ABC$

$$
A=62^{\circ} 20^{\prime} 50^{\prime \prime}, \quad B=55^{\circ} 15^{\prime} 20^{\prime \prime}, \quad a^{2}+b^{2}+c^{2}=m^{2}=225,64 .
$$

Calculate the three sides of the triangle. [^0]


[^0]:    1*) The formulas should be reduced to a form suitable for logarithmic calculation.
Based on well-known trigonometric formulas:

$$
\begin{aligned}
& a=2 R \sin A \\
& b=2 R \sin B \\
& c=2 R \sin C
\end{aligned}
$$

from which

$$
m^{2}=4 R^{2}\left(\sin ^{2} A+\sin ^{2} B+\sin ^{2} C\right)
$$

and

$$
2 R=\frac{m}{\sqrt{\sin ^{2} A+\sin ^{2} B+\sin ^{2} C}}
$$

The latter formula can be transformed into:

$$
2 R=\frac{m}{\sin A \sqrt{1+\frac{\sin ^{2} B}{\sin ^{2} A}+\frac{\sin ^{2} C}{\sin ^{2} A}}}
$$

if now

$$
\frac{\sin B}{\sin A}=\tan \varphi
$$

then

$$
2 R=\frac{m \cos \varphi}{\sin A \sqrt{1+\frac{\sin ^{2} C \cos ^{2} \varphi}{\sin ^{2} A}}}
$$

Furthermore, if

$$
\frac{\sin A}{\cos \varphi \sin C}=\tan \varphi
$$

then

$$
2 R=\frac{m \cos \varphi \sin \varphi}{\sin A}
$$

or, from the 2nd equation,

$$
\frac{\cos \varphi}{\sin A}=\frac{1}{\sin C \tan \varphi}
$$

therefore

$$
2 R=\frac{m \cos \varphi}{\sin C}
$$

The necessary formulas for the calculations are:

$$
\begin{gathered}
\log \tan \varphi=\log \sin B-\log \sin A \\
\log \tan \varphi=\log \sin A-\log \cos \varphi-\log \sin C \\
\log 2 R=\log m-\log \sin C+\log \cos \varphi \\
\log a=\log 2 R+\log \sin A \\
\log b=\log 2 R+\log \sin B \\
\log c=\log 2 R+\log \sin C
\end{gathered}
$$

Auxiliary calculations:

$$
\begin{aligned}
& A+B=120^{\circ} 36^{\prime} 10^{\prime \prime} \\
& C=59^{\circ} 23^{\prime} 50^{\prime \prime} \\
& \log \sin A=9,958493 \\
& \log \sin B=9,914714 \\
& \log \sin C=9,934861 \\
& \hline
\end{aligned}
$$

Final calculations:

Calculation of \(a\):

| 1,001387 |  |
| :---: | :---: |
| $\underline{9,958493}$ |  |
| $\log a=\overline{0,959880}$ |  |
| $\underline{76}$ | 9,1175 |
| 4 | 8 |
| $a=$ | $\overline{9,1175 \delta}$ |

Calculation of \(\varphi\):

| 9,914714 |  |
| ---: | ---: |
| $\log \tan \varphi=\frac{9,958493}{9,956221}$ |  |
| $\frac{15}{6}$ | $42^{\circ} 7^{\prime} 0^{\prime \prime}$ |
| $\varphi=$ | $1^{\prime \prime}$ |

Calculation of \(b\):

| 1,001387 |
| :---: |
| 9,914714 |
| $\log b=0,916101$ |
| $b=\quad 8,24330$ |

Calculation of \(\log \cos \varphi\):

$$
\begin{array}{cr}
42^{\circ} 7^{\prime} 10^{\prime \prime} & 9,870257 \\
-9^{\prime \prime} & 17 \\
\hline \log \cos \varphi=9,870274
\end{array}
$$

## Auxiliary calculations:

Calculation of \(\varphi\):

| 9,958493 |  |
| :---: | ---: |
| $\frac{-9,870274}{c}$ |  |
| 0,088219 |  |
| 9,934861 |  |
| $\log \tan \varphi=$ | 0,153358 |
| $\underline{40}$ | $54^{\circ} 54^{\prime} 40^{\prime \prime}$ |
| 18 | $4^{\prime \prime}$ |
| $\varphi=$ | $54^{\circ} 54^{\prime} 44^{\prime \prime}$ |

Final calculations:

Calculation of \(c\):

![](https://cdn.mathpix.com/cropped/2024_05_02_62e40fe5381661f56debg-2.jpg?height=271&width=423&top_left_y=1838&top_left_x=840)

Calculation of \(\log \cos \varphi\):

$$
\begin{array}{cr}
54^{\circ} 54^{\prime} 50^{\prime \prime} & 9,759522 \\
-6^{\prime \prime} & 18 \\
\hline \log \cos \varphi=9,759540 \\
\hline
\end{array}
$$

Results:

$$
\begin{aligned}
& a=9,11758 \\
& b=8,24330 \\
& c=8,63472 \\
& \hline
\end{aligned}
$$

$2 \log m=2,353416$

$\underline{\log m=1,176708}$

Calculation of \(2 R\):

| 1,176708 |
| ---: |
| $-\quad \frac{9,934861}{1,241847}$ |
| $+\quad 9,759540$ |
| $\log 2 R=1,001387$ |

The problem was solved by: Friedmann Bernát, S.-A.-Ujhely; Grünhut Béla, Pécs; Jankovich György, Losoncz; Jorga Gergely, Gilád; Meitner Elemér, Budapest; Sramkó Loránd, Rimaszombat; Visnya Aladár and Weisz Lipót, Pécs.