Download To solve an oblique triangle means to find the lengths of its sides

If none of the angles of a triangle is a right
angle, the triangle is called oblique.
To solve an oblique triangle means to
find the lengths of its sides and the
measurements of its angles.
All angles are acute
Two acute angles, one obtuse angle
A
FOUR CASES
CASE 1: One side and two angles are
known (SAA or ASA).
CASE 2: Two sides and the angle opposite
one of them are known (SSA).
CASE 3: Two sides and the included
angle are known (SAS).
S
A
ASA
S
A
CASE 4: Three sides are known (SSS).
S
A
SAA
CASE 1: ASA or SAA
S
A
A
S
S
CASE 2: SSA
CASE 3: SAS
1
S
S
S
The Law of Sines is used to solve
triangles in which Case 1 or 2
holds. That is, the Law of Sines
is used to solve SAA, ASA or
SSA triangles.
CASE 4: SSS
Proof (one case)
Theorem Law of Sines
a
b
For a triangle with sides a , b, c and opposite
angles α , β ,γ , respectively,
h
α
sin α sin β sin γ
=
=
a
b
c
sin α =
β
opp h
= , so h = b sin α
hyp b
sin β =
opp h
= , so h = a sin β
hyp a
h = b sin α = a sin β
sin α
sin β
=
a
b
Solve the triangle: α = 30o , β = 70o , a = 5 (SAA)
o
30
α + β + γ = 180o
b
γ
c
α + β + γ = 180o
30o + 70o + γ = 180o
γ = 80o
sin 30o sin 70o sin 30o sin 80o
=
=
70o
b
5
c
5
5
o
o
5 sin 80
5 sin 70
b=
≈ 9.40 c =
≈ 9.85
o
sin 30o
sin 30
2
Solve the triangle: α = 20o , β = 60o , c = 12 (ASA)
20o
α + β + γ = 180o
20o + 60o + γ = 180o
b
12
60o
γ = 100o
γ
o
o
sin 20o sin 100o sin 60 = sin 100
=
12
b
a
12
o
12 sin 60
12 sin 20o
≈ 10.55
a=
≈ 4.17 b =
o
sin 100o
sin 100
a
o
o
sin 132.5 sin 30
=
5
a
a=
5 sin 132.5o
≈ 7.37
sin 30o
a ≈ 7.37, b = 5, c = 3
α ≈ 132.5o , β = 30o ,γ ≈ 17.5o
Solve the triangle: b = 5, c = 3, β = 30o (SSA)
o
30
3
α
a
γ
5
sin 30o sin γ
=
5
3
3 sin 30o
sin γ =
= 0.3
5
γ 1 ≈ 17.5o
γ 2 ≈ 162.5o
β + γ 2 = 30o + 162.5o = 192.5o > 180o
α = 180o − 30o − 17.5o ≈ 132.5o
Solve the triangle: b = 8, c = 10, β = 45o (SSA)
sin 45o sin γ
=
8
10
10 sin 45o
sin γ =
≈ 0.88
8
γ 1 ≈ 62.1o or γ 2 ≈ 117.9o
45o + 62.1o < 180o
45o + 117.9o < 180o
Two triangles!!
Triangle 1: γ 1 ≈ 62.1o
o
Triangle 2: γ 2 ≈ 117.9
α1 = 180o − 45o − 62.1o ≈ 72.9o
α 2 = 180o − 45o − 117.9o ≈ 17.1o
sin 72.9o sin 45o
=
8
a1
sin 17.1o sin 45o
=
8
a2
o
8 sin 17.1
a2 =
≈ 3.33
sin 45o
a1 =
8 sin 72.9
≈ 10.81
sin 45o
a1 ≈ 10.81, b = 8, c = 10
a2 ≈ 3.33, b = 8, c = 10
α1 ≈ 72.9o , β = 45o , γ 1 ≈ 62.1o
α 2 ≈ 17.1o , β = 45o , γ 2 ≈ 117.9o
3
Solve the triangle: c = 5, b = 3, β = 50o (SSA)
sin 50o sin γ
=
3
5
5 sin 50o
sin γ =
3
3
5
50o
a
sin γ ≈ 128
.
No triangle with the given
measurements!
4