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SSA Triangle: The Ambiguous Case

SSA Triangle: The Ambiguous Case
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2012-02-28 10:27
TriangleTrigonometryGeometric Construction
In collections IMP2GBD Triangle Congruence Triangle Congruence Theorems
This example illustrates the SSA case for triangles, where two sides and one of their opposite angles are given, leading to an ambiguous situation. Key mathematical concepts include trigonometry and triangle properties. Using GeoGebra commands such as Circle, Intersect, Segment, and Angle, the construction process is dynamically demonstrated. Suitable for high school precalculus or trigonometry courses.

GeoGebra Commands & Parameters

  1. Circle(C, a04SidePendulum) → d
  2. Intersect(xAxis, yAxis) → A
  3. Segment(A, C) → b
  4. Angle(B_2, C, B_1) → β
  5. If(a04SidePendulum ≥ (b * sin(a06AngleGiven)), β, sweepdefault) → sweep
  6. Segment(C, B) → a
  7. Angle((1, 0), A, C) → α
  8. Polygon(A, C, B_2) → poly2, b_2, a_2, c_2
  9. Polygon(A, C, B_1) → poly1, b_1, a_1, c_1
  10. Corner(5) → dims
  11. If(a07labels ≤ 2, "A", If[a07labels ≤ 4, "B", "C"]) → text1a
  12. If(a06AngleGiven < 90°, If[a04SidePendulum < (a05SideGiven * sin(a06AngleGiven)), 0, (a05SideGiven * sin(a06AngleGiven)) < a04SidePendulum < a05SideGiven, 2, 1], If[a05SideGiven < a04SidePendulum, 1, 0]) → numtriangles
  13. Textfield(a06AngleGiven) → a03InputBox3
  14. Textfield(SidePendulumTemp) → a01InputBox1
  15. Textfield(SideGivenTemp) → a02InputBox2
  16. Max(RemoveUndefined[{x(A), x(B_1), x(B_2), x(C) + abs(((a05SideGiven * sin(a06AngleGiven)) * sin(sweepdefault / 2)))}]) → xmax0
  17. Min(If[Defined[B_1], {x(A), x(B_1), x(B_2), x(C)}, {x(A), x(C), x(C) - abs((a04SidePendulum * sin(sweep / 2)))}]) → xmin0
  18. Max(RemoveUndefined[{y(A), y(B_1), y(B_2), y(C), y(C) - a04SidePendulum}]) → ymax0
  19. Min(If[Defined[B_1], {y(A), y(B_1), y(B_2), y(C), y(C) - a04SidePendulum}, {y(A), y(C), y(C) - a04SidePendulum}]) → ymin0
  20. PolyLine((xmin0, ymin0), (xmax0, ymin0), (xmax0, ymax0), (xmin0, ymax0), (xmin0, ymin0)) → windowframe0
  21. If(xcenterratio ≥ 0.5, xcenter0 - ((xmax0 - xcenter0) * xcenterratio / (1 - xcenterratio)), xmin0) → xmin1
  22. If(xcenterratio ≥ 0.5, xmax0, xcenter0 + ((xcenter0 - xmin0) * (1 - xcenterratio) / xcenterratio)) → xmax1
  23. If(ycenterratio ≥ 0.5, ycenter0 - ((ymax0 - ycenter0) * ycenterratio / (1 - ycenterratio)), ymin0) → ymin1
  24. If(ycenterratio ≥ 0.5, ymax0, ycenter0 + ((ycenter0 - ymin0) * (1 - ycenterratio) / ycenterratio)) → ymax1
  25. PolyLine((xmin1, ymin1), (xmax1, ymin1), (xmax1, ymax1), (xmin1, ymax1), (xmin1, ymin1)) → windowframe1
  26. If(windowratio1 > dimsratio, ymax1, ycenter1 + (dimsratio * (xmax1 - xcenter1))) → ymax2
  27. If(windowratio1 < dimsratio, xmax1, xcenter1 + (ymax1 - ycenter1) / (dimsratio)) → xmax2
  28. PolyLine((xmin2, ymin2), (xmax2, ymin2), (xmax2, ymax2), (xmin2, ymax2), (xmin2, ymin2)) → windowframe2
  29. PolyLine((xmin3, ymin3), (xmax3, ymin3), (xmax3, ymax3), (xmin3, ymax3), (xmin3, ymin3)) → windowframe3
  30. Max(RemoveUndefined[tempL1]) → c
  31. RemoveUndefined({x(A), x(B_1), x(B_2), x(C) + abs((a04SidePendulum * sin(sweepdefault / 2)))}) → tempL2

AI Drawing Prompt Reference

Draw a triangle with two sides a and b, and the angle α opposite side a. According to the SSA condition, there may be 0, 1, or 2 solutions. First, draw side b, then draw a circle with center C and radius a, intersecting the x-axis at point A. Next, draw angle α, connect point A and C to form side b. Depending on the given side a and angle α, there may be 0, 1, or 2 solutions, and mark the possible triangles with different colors.
The prompt above was compiled from the original drawing commands and is for reference only.