Ambiguous case of Sines P1
Montāžas Teksta
- J.J. and Janet went camping in the forest and found comfortable level ground. Janet noticed that J.J.'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
- Janet knew he was wrong and she was going to teach him about the AMBIGUOUS CASE OF LAW OF SINES
- ° ∠
-
- 180 - (64.6+60) = 55.4 C1 = 55.4°
- Set up Law of Sines, cross multiply, and find arcsin of ∠B to get ∠B1 = 64.6°.
- Add angles A & B1 and subtract that from 180° to find ∠C1
- 7 7.3sin(60) sin(B)
- 7 csin(60) sin(55.4)
- Set up Law of Sines including ∠C1 & c1, cross multiply, and get c1 alone so c1 = 6.7 ft
- =
- =
- 7sin(55.4)sin(60)
- B1 = sin
- -1
- = 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 J.J.'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 180 - 64.6 = 115.4 so ∠B2 = 115.4°
- B2
- C
- C
- C
- B1
- A= 60°a = 7b=7.3
-
- ∠B1 = 64.6° ∠C1 = 55.4° c1 = 6.7 ft∠B2 = 115.4° ∠C2 = 4.6° c2 = 0.6 ft
- 7 csin(60) sin(4.6)
- 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
- =
- 7sin(4.6)sin(60)
- =
- c2
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