Calc Final 1

LES FOR MEG

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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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