Calculate the approximate angular height of the Sun at the upper climax on the day of the Olympiad in Kazan (φ = 55˚47´).

The day of the Olympiad practically coincides with the day of the autumnal equinox, so the declination of the Sun can be taken equal to 00. According to the formula hv.k. = (90-φ) + δ, we find hv.k. = (900-55047´) = 34˚13´

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