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Section 6.1 - SOLVING OBLIQUE TRIANGLES
An oblique triangle is a triangle that does not have a right angle.
To solve for any missing angles or sides, we cannot use the
Pythagorean Theorem or our typical sine, cosine, or tangent
function ratios.
SOH-CAH-TOA does NOT work on oblique triangles!
The Pythagorean Theorem does NOT work on oblique triangles!!!
We solve these triangles by use the Law of Sines or the Law of
Cosines. The information given will determine which “Law” we
use.
THE LAW OF SINES
Use the Law of Sines when the information marked in red
is given.
a
b
c
Law of Sine :
=
=
sin A sin B sin C
ASA
2 angles and any
side
AAS
2 angles and any side
SSA
2 sides and a nonincluded
angle (ambiguous case)
If the known parts of the triangle are ASA, AAS,
or SSA, use the Law of Sines. The known parts
are shown in red in the triangles above.
Law of Sines
sin A sin B sin C


a
b
c
or
a
b
c


sin A sin B sin C
The sine law (rule) for any arbitrarily
shaped triangle:
A
b
c
a
b
c
law of sines :
=
=
sinA sinB sinC
C
a
B
In order to use the law of sines, you must know
a) two angles and any side (AAS / ASA) or
b) two sides and a angle opposite one of them (SSA),
the ambiguous case.
Example 1: Solve the triangle if C = 28.00, c = 46.8cm,
(AAS)
and B = 101.50
Solution: Draw and label the triangle.
The given information is in red. This
is AAS. Therefore use the Law of
Sines.
B
101.5⁰
28.0⁰
A
C
c
b

sin C sin B
46.8cm
b

0
sin 28.0
sin101.50
b(sin28.0 )  (sin 101.5 )(46.8 cm)
0
0
(sin 101.50 )(46.8 cm)
b
 97.7 cm
0
sin 28.0
(Continued…)
A = 1800 – B – C = 1800 -101.50 – 28.00 = 50.50
To find side a,
c
a

sin C sin A
46.8cm
a

0
sin28.0 sin50.50
a(sin28.0 )  (sin50.5 )(46.8cm)
0
0
(sin50.50 )(46.8cm)
a
(sin28.00 )
 76.90
The solution is a = 76.9 cm, b = 97.7 cm, and A = 50.5⁰
Example 2 - ASA
Solve the triangle below.
C
b
a
37 
18 
21.7
Ambiguous Case
When the given information is SSA, there are three
possibilities.
1) (1 triangle) - There is only one triangle possible for
the given information.
2) (2 triangles) - There are 2 triangles possible for the
given information.
3) (no triangles) - There is not a triangle possible for the
given information.
Example 2: Solve the triangle: a = 6, b = 8 , A = 35⁰. Round all
answers to the nearest tenth.
B
c
A
6
35⁰
8
a
b

sin A sin B
C
Solution: The information gives a SSA
triangle. Use the Law of Sine. This is
the ambiguous case. There may be 0,
1, or 2 solutions.
6
8

sin35 sin B
B = 49.9⁰ or maybe B =180⁰ - 49.9⁰ = 130.1⁰
Since 130.1⁰ will work in this triangle, there are 2 possible
solutions for this problem. (continued)
Case 1: B1 ≈ 49.9⁰. Then C1 = 180⁰ - 35⁰ - 49.9 ≈ 95.1⁰
c1
6

sin35 sin 95.1
c1 ≈ 10.419
Case 2: B2 ≈ 130.1⁰. Then C2 = 180⁰ - 35⁰ - 130.1⁰ ≈ 14.9⁰
c2
6

sin35 sin14.9
Triangle 1:
B = 49.9, C = 95.1⁰ c ≈ 10.4
c2 ≈ 2.69
Or Triangle 2:
B = 130.1⁰ , C = 14.9⁰, c = 2.7
Example 3 - 2 triangles
Solve the triangle with A = 29.5, a = 14.7, b = 20.1
C
20.1
14.7

A
29.5
c
B
Example 4: Solve the triangle where a = 2, c = 1, and C = 50⁰
This information gives SSA, the ambiguous case of the Law of Sine.
a
c

sin A sin C
1
2

sin A sin 50
Solving for A gives sinA ≈ 1.53.
Since the sine function cannot be greater than 1, there is no
solution for this problem.
Example 5 SSA – One Solution
Solve Triangle ABC if A = 43 degrees, a = 81, and
b = 62. Round lengths of sides to the nearest tenth
and angle measures to the nearest degree.
Finding Area of an Oblique Triangle
Area 
1
bc sin A
2
1
 ac sin B
2

The area of a triangle equals ½
the product of the lengths of
two sides times the sine of
their included angle.
1
ab sin C
2
Example 6:
Find the area of a triangle having two sides of lengths 8 meters and
12 meters and an included angle of 135⁰. Round to the nearest
square meter.
Area = (1/2)(8)(12)sin135⁰ ≈ 34 m2
Example 7
Find the area of the given triangle
56.1

63
82.6
Example 8
The figure shows a 1200-yard-long sand beach and an oil
platform in the ocean. The angle made with the platform
from one end of the beach is 85⁰ and from the other end is
76⁰. Find the distance of the oil platform, to the nearest
tenth of a yard, from each end of the beach.
76⁰
85⁰