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SSA Ambiguous Case: Dynamic Triangle Solutions

SSA Ambiguous Case: Dynamic Triangle Solutions
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2020-11-13 03:35
TriangleAngleTrigonometryGeometric ConstructionSlider Animation
In collections Triangle Congruency Exploration
① This example visually demonstrates the ambiguous case of Side-Side-Angle (SSA) in triangle construction, where given two sides and a non-included angle, the triangle may have zero, one, or two possible solutions. ② Core mathematical concepts include the sine rule, triangle angle sum, and the non-uniqueness of SSA as a congruence criterion)Skip. ③ Construction highlights: using GeoGebra commands Circle, Point, Segment, Rotate, Angle, Ray, and Intersect, a circle is drawn, a side is rotated, and intersections with a ray are found, dynamically showing two possible vertices (B' and F') as parameters change. ④ This is ideal for high school or early college students studying trigonometry, triangle solving, and geometric dynamic exploration, and for teachers demonstrating the concept in class.

GeoGebra Commands & Parameters

  1. Circle(H, 6) → c
  2. Point(c) → Y
  3. Segment(H, Y) → f
  4. Circle(Y, 5) → d
  5. Point(d) → C
  6. Segment(Y, C) → g
  7. Rotate(Y, 50°, H) → B'
  8. Angle(Y, H, B') → α
  9. Ray(H, B') → h
  10. Point(h) → D
  11. Segment(H, D) → i
  12. Circle(S, 6) → e
  13. Point(e) → P
  14. Segment(S, P) → j
  15. Circle(P, 5) → k
  16. Rotate(P, 50°, S) → F'
  17. Angle(P, S, F') → β
  18. Ray(S, F') → l
  19. Intersect(k, l, 1) → U
  20. Segment(P, U) → m
  21. Segment(S, U) → n

AI Drawing Prompt Reference

Draw a triangle with two sides of lengths 6 and 5, and an included angle of 50 degrees. First, draw a circle with center H and radius 6, then draw another circle with center Y (a point on the first circle) and radius 5. One intersection of the two circles is point C. Next, rotate point Y around H by 50 degrees to get point B', and draw a ray from H to B'. On this ray, take a point D and connect HD. Then, draw a circle with center P and radius 6, and rotate point P around S by 50 degrees to get point F', and draw a ray from S to F'. The intersection of the two rays is point U, and connect PU and SU to complete the triangle.
The prompt above was compiled from the original drawing commands and is for reference only.