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  • John and Emma went camping in the forest and found comfortable level ground. Emma noticed that John's tent entrance looked odd.
  • 7.3sin(60)7
  • 
  • A
  • My 7-foot pole, 7.3-foot pole, and desired 60° angle only make this triangle for the opening of the tent!
  • B
  • C
  • A= 60° a = 7b=7.3
  • Emma knew he was wrong and she was going to teach him about the AMBIGUOUS CASE OF LAW OF SINES
  • 
  • Set up Law of Sines, cross multiply, and find arcsin of ∠B to get ∠B1 = 64.6°.
  • 7sin(60)
  • Add angles A & B1 and subtract that from 180° to find ∠C1
  • 180 - (64.6+60) = 55.4 
  • 7sin(60)
  • Set up Law of Sines including ∠C1 & c1, cross multiply, and get c1 alone so c1 = 6.7 ft
  • =
  • =
  • 7.3sin(B)
  • csin(55.4)
  • 7sin(55.4)sin(60)
  • B1 = sin
  • -1
  • C1 = 55.4°
  • = c1
  • Think of swinging side "a" on a hinge until it makes another triangle. Because this would make an isosceles triangle, the two angles would be equal. A line measures 180° and so taking B1 from it gives the supplementary angle. This would be ∠B2. This gives us John's triangle.
  • *In other cases, add the B2 with ∠A to make sure it is less than 180, if not, there is no other triangle.*
  • A
  • 180 - B1 = B2 
  • B2
  • C
  • 180 - 64.6 = 115.4
  • C
  • so ∠B2 = 115.4°
  • C
  • B1
  • A= 60°a = 7b=7.3
  •  
  • 7sin(60)
  • ∠B1 = 64.6° ∠C1 = 55.4° c1 = 6.7 ft∠B2 = 115.4° ∠C2 = 4.6° c2 = 0.6 ft
  • Set up Law of Sines including  ∠C2 & c2, cross multiply, and get c2 alone so c2 = 0.6 ft
  • Take the known angles of the second triangle from 180 to find the last angle, ∠C2
  • 180 - (115.4+60) = 4.6 so ∠C2 = 4.6 
  • csin(4.6)
  • =
  • 7sin(4.6)sin(60)
  • =
  • c2
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