If none of the angles of a triangle is a right angle, the triangle is called oblique. All angles are acute Two acute angles, one obtuse angle.

If none of the angles of a triangle is a right angle, the triangle is called oblique.

All angles are acute

Two acute angles, one obtuse angle

To solve an oblique triangle means to find the lengths of its sides and the measurements of its angles.

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).

CASE 4: Three sides are known (SSS).

CASE 1: ASA or SAA

S

A

AASA

SA A

SAA

S

SA

CASE 2: SSA

S

SA

CASE 3: SAS

S

S

S

CASE 4: SSS

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.

Law of Sines

For a triangle with sides and opposite

angles , , , respectively,

a b c, ,

sin sin sin a b c

Law of Sines

c

C

b

B

a

A sinsinsin

For a triangle with sides a, b, and c, and angles A, B, and C

Solve the triangle: = 30 (SAA) , , 70 5a

70

30

5

b c

180

30 70 180

80

sin sin305

70

b

b 5 70

309 40

sin

sin.

sin sin305

80

c

c 5 80

309 85

sin

sin.

Solve the triangle: = 20 (ASA) , , 60 12c

20

60

12

a

b

180

20 60 180 100

sin sin20 10012

a

a 12 20

1004 17

sin

sin.

sin sin60 10012

b

b 12 60

10010 55

sin

sin.

The area A of a triangle is

A bh12

where b is the base and h is the altitude drawn to that base.

h

b

a

sin ha

h a sin

A bh b a 12

12

sin 12

absin

The area A of a triangle equals one-half the product of two of its sides times the sine of its included angle.

A ab

A ac

A bc

121212

sin

sin

sin

Find the area of the triangle for which A a

c

5

7 70

,

, .

A ac12

sin

12

5 7 70sin

16 44.

Find the area of a triangle ABC if a = 5, C = 65 degrees,

and B = 45 degrees.

Solve the triangle: (SSA)c b 5 3 50, ,

sin sin503 5

sinsin 5 50

3

sin . 128

No triangle with the given measurements!

3

5

50a

Solve the triangle: (SSA)b c 5 3 30, ,

30

5

3

a

sin sin305 3

sinsin

. 3 305

0 3

1 17 5 . 2 162 5 .

2 30 162 5 192 5 180 . .

180 30 17 5 132 5 . .

sin . sin132 5 305

a

a 5 132 5

307 37

sin .

sin.

a b c

7 37 5 3

132 5 30 17 5

. , ,

. , , .

Solve the triangle: (SSA)b c 8 10 45, ,

sin sin458 10

sinsin

. 10 458

0 88

1 262 1 117 9 . . or

45 62 1 180 . 45 117 9 180 .

Two triangles!!

Triangle 1: 1 62 1 .

1 180 45 62 1 72 9 . .

sin . sin72 9 4581

a

a18 72 9

4510 81 sin .

sin.

a b c1

1 1

10 81 8 10

72 9 45 62 1

. , ,

. , , .

Triangle 2: 2 117 9 .

2 180 45 117 9 17 1 . .

sin . sin17 1 4582

a

a28 17 1

453 33 sin .

sin.

a b c2

2 2

3 33 8 10

17 1 45 117 9

. , ,

. , , .

Lesson Overview 5-6B

5-Minute Check Lesson 5-7A

Lesson Overview 5-7A

Lesson Overview 5-7B

5-Minute Check Lesson 5-8A

Heron’s Formula

The area A of a triangle with sides a, b, and c is

A s s a s b s c

where s a b c 12

.

Find the area of a triangle whose sides are

5, 8, and 11.

Let a b c 5 8 11, ,

s a b c 12

12

5 8 11 12

A s s a s b s c

12 12 5 12 8 12 11

336 4 21

Lesson Overview 5-6A