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In 2015, an anonymous Japanese woman using the pen name "aerile re" published the first known method (the method of 3 circumcenters) to construct a proof in elementary geometry for a special class of adventitious quadrangles problem. [7] [8] [9] This work solves the first of the three unsolved problems listed by Rigby in his 1978 paper. [5]
Since there are four such triangles, there are four such constraints on sums of angles, and the number of degrees of freedom is thereby reduced from 12 to 8. The four relations given by the sine law further reduce the number of degrees of freedom, from 8 down to not 4 but 5, since the fourth constraint is not independent of the first three ...
The extended base of a triangle (a particular case of an extended side) is the line that contains the base. When the triangle is obtuse and the base is chosen to be one of the sides adjacent to the obtuse angle, then the altitude dropped perpendicularly from the apex to the base intersects the extended base outside of the triangle. The area of ...
The base of the triangle is formed by the distance between two measuring stations and the angle θ is the parallax angle formed by the object as seen by the two stations. This baseline is usually very long for best accuracy; in principle the stations could be on opposite sides of the Earth. However, this distance is still short compared to the ...
Since no triangle can have two obtuse angles, γ is an acute angle and the solution γ = arcsin D is unique. If b < c, the angle γ may be acute: γ = arcsin D or obtuse: γ ′ = 180° − γ. The figure on right shows the point C, the side b and the angle γ as the first solution, and the point C ′, side b ′ and the angle γ ′ as the ...
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The red section on the right, d, is the difference between the lengths of the hypotenuse, H, and the adjacent side, A.As is shown, H and A are almost the same length, meaning cos θ is close to 1 and θ 2 / 2 helps trim the red away.
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