For perimeter $15$, $\angle B = 100^{\circ}, \angle C = 90^{\circ}$, the construction is

For perimeter $15$, $\angle B = 100^{\circ}, \angle C = 90^{\circ}$, the construction is
  1. impossible (sum of base angles > $180^{\circ}$)
  2. possible
  3. requires angle bisector
  4. yields infinite triangles

Solution

$100 + 90 = 190 > 180$, no valid triangle.

Asked in: MH-SSC-9

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