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.

Section 6.1 - SOLVING OBLIQUE TRIANGLES

ASA2 angles and any side

AAS2 angles and any side

SSA 2 sides and a nonincluded angle (ambiguous case)

THE LAW OF SINES

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.

Use the Law of Sines when the information marked in red is given.

sin sin sin

a b c

A B CLaw of Sine: = =

Law of Sines

or

c

Csin

b

Bsin

a

Asin

Csin

c

Bsin

b

Asin

a

cb

a

a b claw of sines: = =

sinA sinB sinC

The sine law (rule) for any arbitrarily shaped triangle:

C

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.

A C

Example 1: Solve the triangle if C = 28.00, c = 46.8cm, (AAS) and B = 101.50

sin sin

c b

C B

0 0

46.8

sin 28.0 sin101.5

cm b

0 0(sin28.0 ) (sin 101.5 )(46.8 )b cm0

0

(sin 101.5 )(46.8 )

sin 28.7 7 c

09 . m

cmb

BSolution: Draw and label the triangle. The given information is in red. This is AAS. Therefore use the Law of Sines.

101.5⁰

28.0⁰

c = 46.8

(Continued…)

A = 1800 – B – C = 1800 -101.50 – 28.00 = 50.50

To find side a,

sin sin

c a

C A

0 0

46.8

sin28.0 sin50.5

cm a

0 0(sin28.0 ) (sin50.5 )(46.8 )a cm0

0

(sin50.5 )(46.8 )

(sin28.0 )

cma

076.9

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.

A8

6

C

B

35⁰

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.

sin sin

a b

A B 6 8

sin35 sin B

B = 49.9⁰ or maybe B =180⁰ - 49.9⁰ = 130.1⁰Since 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

16

sin35 sin 95.1

c c1 ≈ 10.419

Case 2: B2 ≈ 130.1⁰. Then C2 = 180⁰ - 35⁰ - 130.1⁰ ≈ 14.9⁰

26

sin35 sin14.9

c c2 ≈ 2.69

Triangle 1:B = 49.9, C = 95.1⁰

c ≈ 10.4 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

29.5 A 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.

sin sin

a c

A C 12

sin sin 50A

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 Triangle1

Area sin2

1sin

2

1sin

2

bc A

ac B

ab C

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

The area of a triangle equals ½ the product of the lengths of two sides times the sine of their included angle.

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.

85⁰

76⁰

1200 yds