C
C
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 c
sin(60) sin(4.6)
A= 60°
a = 7
b=7.3
My 7-foot pole, 7.3-foot pole, and desired 60° angle only make this triangle for the opening of the tent!
7 7.3
sin(60) sin(B)
Janet knew he was wrong and she was going to teach him about the AMBIGUOUS CASE OF LAW OF SINES
=
A
( )
B
C
7.3sin(60)
7
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
180 - (64.6+60) = 55.4 C1 = 55.4°
7 c
sin(60) sin(55.4)
=
=
° ∠
B1 = sin
-1
7sin(55.4)
sin(60)
= c1
Set up Law of Sines including ∠C1 & c1, cross multiply, and get c1 alone so c1 = 6.7 ft
A
180 - B1 = B2 180 - 64.6 = 115.4 so ∠B2 = 115.4°
C
B2
B1
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.*
180 - (115.4+60) = 4.6 so ∠C2 = 4.6
Take the known angles of the second triangle from 180 to find the last angle, ∠C2
A= 60°
a = 7
b=7.3
7sin(4.6)
sin(60)
=
c2
Set up Law of Sines including ∠C2 & c2, cross multiply, and get c2 alone so c2 = 0.6 ft
∠B1 = 64.6° ∠C1 = 55.4° c1 = 6.7 ft
∠B2 = 115.4° ∠C2 = 4.6° c2 = 0.6 ft
C
C
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 c
sin(60) sin(4.6)
A= 60°
a = 7
b=7.3
My 7-foot pole, 7.3-foot pole, and desired 60° angle only make this triangle for the opening of the tent!
7 7.3
sin(60) sin(B)
Janet knew he was wrong and she was going to teach him about the AMBIGUOUS CASE OF LAW OF SINES
=
A
( )
B
C
7.3sin(60)
7
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
180 - (64.6+60) = 55.4 C1 = 55.4°
7 c
sin(60) sin(55.4)
=
=
° ∠
B1 = sin
-1
7sin(55.4)
sin(60)
= c1
Set up Law of Sines including ∠C1 & c1, cross multiply, and get c1 alone so c1 = 6.7 ft
A
180 - B1 = B2 180 - 64.6 = 115.4 so ∠B2 = 115.4°
C
B2
B1
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.*
180 - (115.4+60) = 4.6 so ∠C2 = 4.6
Take the known angles of the second triangle from 180 to find the last angle, ∠C2
A= 60°
a = 7
b=7.3
7sin(4.6)
sin(60)
=
c2
Set up Law of Sines including ∠C2 & c2, cross multiply, and get c2 alone so c2 = 0.6 ft
∠B1 = 64.6° ∠C1 = 55.4° c1 = 6.7 ft
∠B2 = 115.4° ∠C2 = 4.6° c2 = 0.6 ft
C
C
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 c
sin(60) sin(4.6)
A= 60°
a = 7
b=7.3
My 7-foot pole, 7.3-foot pole, and desired 60° angle only make this triangle for the opening of the tent!
7 7.3
sin(60) sin(B)
Janet knew he was wrong and she was going to teach him about the AMBIGUOUS CASE OF LAW OF SINES
=
A
( )
B
C
7.3sin(60)
7
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
180 - (64.6+60) = 55.4 C1 = 55.4°
7 c
sin(60) sin(55.4)
=
=
° ∠
B1 = sin
-1
7sin(55.4)
sin(60)
= c1
Set up Law of Sines including ∠C1 & c1, cross multiply, and get c1 alone so c1 = 6.7 ft
A
180 - B1 = B2 180 - 64.6 = 115.4 so ∠B2 = 115.4°
C
B2
B1
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.*
180 - (115.4+60) = 4.6 so ∠C2 = 4.6
Take the known angles of the second triangle from 180 to find the last angle, ∠C2
A= 60°
a = 7
b=7.3
7sin(4.6)
sin(60)
=
c2
Set up Law of Sines including ∠C2 & c2, cross multiply, and get c2 alone so c2 = 0.6 ft
∠B1 = 64.6° ∠C1 = 55.4° c1 = 6.7 ft
∠B2 = 115.4° ∠C2 = 4.6° c2 = 0.6 ft
C
C
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 c
sin(60) sin(4.6)
A= 60°
a = 7
b=7.3
My 7-foot pole, 7.3-foot pole, and desired 60° angle only make this triangle for the opening of the tent!
7 7.3
sin(60) sin(B)
Janet knew he was wrong and she was going to teach him about the AMBIGUOUS CASE OF LAW OF SINES
=
A
( )
B
C
7.3sin(60)
7
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
180 - (64.6+60) = 55.4 C1 = 55.4°
7 c
sin(60) sin(55.4)
=
=
° ∠
B1 = sin
-1
7sin(55.4)
sin(60)
= c1
Set up Law of Sines including ∠C1 & c1, cross multiply, and get c1 alone so c1 = 6.7 ft
A
180 - B1 = B2 180 - 64.6 = 115.4 so ∠B2 = 115.4°
C
B2
B1
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.*
180 - (115.4+60) = 4.6 so ∠C2 = 4.6
Take the known angles of the second triangle from 180 to find the last angle, ∠C2
A= 60°
a = 7
b=7.3
7sin(4.6)
sin(60)
=
c2
Set up Law of Sines including ∠C2 & c2, cross multiply, and get c2 alone so c2 = 0.6 ft
∠B1 = 64.6° ∠C1 = 55.4° c1 = 6.7 ft
∠B2 = 115.4° ∠C2 = 4.6° c2 = 0.6 ft
C
C
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 c
sin(60) sin(4.6)
A= 60°
a = 7
b=7.3
My 7-foot pole, 7.3-foot pole, and desired 60° angle only make this triangle for the opening of the tent!
7 7.3
sin(60) sin(B)
Janet knew he was wrong and she was going to teach him about the AMBIGUOUS CASE OF LAW OF SINES
=
A
( )
B
C
7.3sin(60)
7
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
180 - (64.6+60) = 55.4 C1 = 55.4°
7 c
sin(60) sin(55.4)
=
=
° ∠
B1 = sin
-1
7sin(55.4)
sin(60)
= c1
Set up Law of Sines including ∠C1 & c1, cross multiply, and get c1 alone so c1 = 6.7 ft
A
180 - B1 = B2 180 - 64.6 = 115.4 so ∠B2 = 115.4°
C
B2
B1
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.*
180 - (115.4+60) = 4.6 so ∠C2 = 4.6
Take the known angles of the second triangle from 180 to find the last angle, ∠C2
A= 60°
a = 7
b=7.3
7sin(4.6)
sin(60)
=
c2
Set up Law of Sines including ∠C2 & c2, cross multiply, and get c2 alone so c2 = 0.6 ft
∠B1 = 64.6° ∠C1 = 55.4° c1 = 6.7 ft
∠B2 = 115.4° ∠C2 = 4.6° c2 = 0.6 ft
C
C
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 c
sin(60) sin(4.6)
A= 60°
a = 7
b=7.3
My 7-foot pole, 7.3-foot pole, and desired 60° angle only make this triangle for the opening of the tent!
7 7.3
sin(60) sin(B)
Janet knew he was wrong and she was going to teach him about the AMBIGUOUS CASE OF LAW OF SINES
=
A
( )
B
C
7.3sin(60)
7
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
180 - (64.6+60) = 55.4 C1 = 55.4°
7 c
sin(60) sin(55.4)
=
=
° ∠
B1 = sin
-1
7sin(55.4)
sin(60)
= c1
Set up Law of Sines including ∠C1 & c1, cross multiply, and get c1 alone so c1 = 6.7 ft
A
180 - B1 = B2 180 - 64.6 = 115.4 so ∠B2 = 115.4°
C
B2
B1
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.*
180 - (115.4+60) = 4.6 so ∠C2 = 4.6
Take the known angles of the second triangle from 180 to find the last angle, ∠C2
A= 60°
a = 7
b=7.3
7sin(4.6)
sin(60)
=
c2
Set up Law of Sines including ∠C2 & c2, cross multiply, and get c2 alone so c2 = 0.6 ft
∠B1 = 64.6° ∠C1 = 55.4° c1 = 6.7 ft
∠B2 = 115.4° ∠C2 = 4.6° c2 = 0.6 ft
C
C
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 c
sin(60) sin(4.6)
A= 60°
a = 7
b=7.3
My 7-foot pole, 7.3-foot pole, and desired 60° angle only make this triangle for the opening of the tent!
7 7.3
sin(60) sin(B)
Janet knew he was wrong and she was going to teach him about the AMBIGUOUS CASE OF LAW OF SINES
=
A
( )
B
C
7.3sin(60)
7
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
180 - (64.6+60) = 55.4 C1 = 55.4°
7 c
sin(60) sin(55.4)
=
=
° ∠
B1 = sin
-1
7sin(55.4)
sin(60)
= c1
Set up Law of Sines including ∠C1 & c1, cross multiply, and get c1 alone so c1 = 6.7 ft
A
180 - B1 = B2 180 - 64.6 = 115.4 so ∠B2 = 115.4°
C
B2
B1
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.*
180 - (115.4+60) = 4.6 so ∠C2 = 4.6
Take the known angles of the second triangle from 180 to find the last angle, ∠C2
A= 60°
a = 7
b=7.3
7sin(4.6)
sin(60)
=
c2
Set up Law of Sines including ∠C2 & c2, cross multiply, and get c2 alone so c2 = 0.6 ft
∠B1 = 64.6° ∠C1 = 55.4° c1 = 6.7 ft
∠B2 = 115.4° ∠C2 = 4.6° c2 = 0.6 ft