By Kim B. M.

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**Extra resources for 2-Universal Positive Definite Integral Quinary Diagonal Quadratic Forms**

**Example text**

Therefore BGD, the angle of the greatest southern latitude, is given as an exterior angle of triangle EGD. In accordance with the theorems on Plane Triangles, the opposite interior angle GED will also be given as the angle of the eccentric's maximum southern inclination to the plane of the ecliptic. By means of the minimum southern latitude we shall likewise demonstrate the mm inclination, for example, by mean of angle EFD. In triangle EFD, the ratio of sides EF : ED is given as well as angle EFD.

Let a plane perpendicular to it and passing through its center intersect it in ABC. Let (the ecliptic's] intersection with Venus, orbital plane be DBE. Let the earth's center be A; the center of the a planet's orbit, B; and the angle of the orbit's inclination to the ecliptic, ABE. With B as center, describe orbit DFEG. Draw diameter FBG perpendicular to diameter DE. Let the orbit's plane be conceived to be so related to the assumed perpendicular plane that lines drawn therein perpendicular to DE are parallel to one another and to the plane of the ecliptic, in which FBG is the only [such perpendicular].

Hence, angle MAH is obtained = 1° 47' = the latitudinal declination. But if it is not boring to consider what variation in longitude is produced by this declination of Venus, let us take triangle ALH, understanding LH to be a diagonal of parallelogram LKHM = 5091p where of AL = 4919p. ALH is a right angle. From this information hypotenuse AH is obtained = 7079p. Hence, the ratio of the sides being given, angle HAL = 45° 59'. But MAL was shown = 45° 57'. Therefore, the excess is only 2'. Q. E. D.

### 2-Universal Positive Definite Integral Quinary Diagonal Quadratic Forms by Kim B. M.

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