On which latitude circle of our globe will a degree of longitude be $30 \mathrm{~km}$ ? (The radius of the Earth is $6377 \mathrm{~km}$. - Using logarithm tables!)
We consider the Earth as a sphere. The geographic latitude of a point $P$ is understood as the angle enclosed by the radius of the sphere at that point with the plane of the equator, which can also be measured on the meridian of the point in degrees. (The plane of the meridian passes through the Earth's axis, thus it is perpendicular to the plane of the equator.)

Points at latitude $\varphi$ lie on the parallel of latitude $\varphi$, whose plane is parallel to the plane of the equator. The center of the sphere $O$, the center of the parallel of latitude $K$, and any point $P$ on the parallel of latitude form a right triangle in the plane of the meridian of the latter point (see the figure). The hypotenuse $OP = r$ is the radius of the sphere, the leg $KP$ is the radius $\varrho$ of the parallel of latitude, and the angle at point $P$ is $\varphi$.

![](https://cdn.mathpix.com/cropped/2024_05_02_58b05335a42b3ee0ef87g-1.jpg?height=509&width=501&top_left_y=434&top_left_x=801)

On a parallel of latitude, one degree of longitude is the 360th part of the circumference of the circle with radius $\varrho$: $d = \frac{2 \varrho \pi}{360}$. Since the radius of the parallel of latitude $\varphi$ is $\varrho = r \cos \varphi$, the length of one degree of longitude on this parallel is

$$
d = \frac{2 r \pi \cos \varphi}{360}. \quad \text{Thus} \quad \cos \varphi = \frac{360 d}{2 r \pi}
$$

According to the problem, we need to calculate the $\varphi$ corresponding to $d = 30 \text{ km}$. Substituting the values,

$$
\cos \varphi = \frac{360 \cdot 30}{2 \cdot 6377 \pi} = \frac{5400}{6377 \pi}
$$

Thus,

$$
\lg \cos \varphi = \lg 5400 - (\lg 6377 + \lg \pi) = 9.4307 - 10
$$

and therefore,

$$
\varphi = 74^\circ 22'
$$

Ádám Antal (Bp. VIII., Széchenyi g. III. o. t.)