Module 8Lesson 5

Oblique Triangles

Florben G. Mendoza

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

Florben G. Mendoza

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

sides and the measurements of its angles.

a b

cAB

C

Sides: a

b

c

Angles: A

B

C

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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).

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CASE 1: ASA or SAA

S

A

A

ASA

SA A

SAA

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S

SA

CASE 2: SSA

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S

SA

CASE 3: SAS

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S

S

S

CASE 4: SSS

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Practice Exercise 1:

1)

2)

3)

4)

5)

6)

7)

8)

9)

SAS

SAS

SSA

SSAASA

SAA

SAA

SSS

ASA

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Practice Exercise 2:

1. (A, B, c)

2. (A, B, a)

3. (b, c, A)

4. (a, b, A)

5. (a, b, c)

6. (C, b, c)

7. (a, B, C)

8. (a, A, C)

9. (A, b, C)

10. (C, b, a)

SAA

ASA

SAS

SSA

SSS

SSA

ASA

SAA

ASA

SAS

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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.

ASA

A

AS

SAA

S

A A

SSAS

A

S

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Law of Sines

AB

C

ab

c

Let’s drop an altitude

and call it h.

h

If we think of h as

being opposite to

both A and B, then

sin sinh h

A and Bb a

Let’s solve both for h.

sin sinh b A and h a B

This meanssin sin and dividing by .

sinA sin

a

b A a B ab

B

b

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A B

C

ab

c

If I were to drop an altitude to

side a, I could come up with

sin sinB C

b c

Putting it all together gives us

the Law of Sines.

sin sin sinA B C

a b c

You can also use it

upside-down. sin sin sin

a b c

A B C

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Example 1:45 , 50 , 30Let A B a

A B

C

ab

c45° 50°

= 30

= 180° - (45° + 50°)

Step 1: C = 180° - (A + B)

C = 85°

= 180° - 95°

Step 2: a

sin A=

b

sin B

30

sin 45°=

b

sin 50°

b (sin 45°) = 30 (sin 50°)

sin 45° sin 45°

b = 32.50

SAA

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Example 1: SAA

A B

C

ab

c

45 , 50 , 30Let A B a

45° 50°

= 30 Step 3: a

sin A=

c

sin C

30

sin 45°=

c

sin 85°

c (sin 45°) = 30 (sin 85°)

sin 45° sin 45°

c = 42.26

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Example 2:

Let C = 35°, B = 10°, and a = 45

Step 1: A = 180° - (B + C)

= 180° - (10° + 35°)

= 180° - 45°

A = 135°

A B

ab

c

35°

10°

= 45

CStep 2:

a

sin A=

b

sin B

45

sin 135°=

b

sin 10°

b (sin 135°) = 45 (sin 10°)

sin 135° sin 135°

b = 11.05

ASA

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Example 2: ASA

Let C = 35°, B = 10°, and a = 45

A B

ab

c

35°

10°

= 45

CStep 3:

a

sin A=

c

sin C

45

sin 135°=

c

sin 35°

c (sin 135°) = 45 (sin 35°)

sin 135° sin 135°

c = 36.50

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Ambiguous Case (SSA)

Case 1: If A is acute and a < b

A

C

B

ba

c

h = b sin A

a. If a < b sinA

A

C

B

b

a

c

h

NO SOLUTION

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Case 1: If A is acute and a < b

A

C

B

b a

c

h = b sin A

b. If a = b sinA

A

C

B

b= a

c

h

1 SOLUTION

Ambiguous Case (SSA)

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Case 1: If A is acute and a < b

A

C

B

b a

c

h = b sin A

b. If a > b sinA

A

C

B

b

c

h

2 SOLUTIONS

a a

B

180 -

Ambiguous Case (SSA)

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Case 2: If A is obtuse and a > bC

A B

a

b

c

ONE SOLUTION

Ambiguous Case (SSA)

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Case 2: If A is obtuse and a ≤ bC

A B

a

b

c

NO SOLUTION

Ambiguous Case (SSA)

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Let A = 40°, b = 10, and a = 9

Example 3:

A B

C

ab

c

h= 10 = 9

40°

Step 1: Solve for h

h = b sin A

h = 10 sin 40°

h = 6.43

a > h ( 2 Solutions)

Step 2:a

sin A=

b

sin B

9

sin 40°=

10

sin B

9 (sin B) = 10 (sin 40°)

9 9

sin B = 0.71

B = sin-1 0.71

B = 45.23°

SSA

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Let A = 40°, b = 10, and a = 9

Example 3: SSA

A B

C

ab

c

h= 10 = 9

40°

Step 3: C = 180° - (A + B)

C = 180° - (40° + 45.23°)

C = 180° - 85.23°

C = 94.77°

a

sin A=

c

sin C

9

sin 40°=

c

sin 94.77°

c (sin 40°) = 9 (sin 94.77°)

c = 13.95

Step 4:

(sin 40°) (sin 40°)

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Let A = 40°, b = 10, and a = 9

Example 3: SSA

A B

C

ab

c

h= 10 = 9

40° 40°

b = 10 a = 9

45.23°45.23°BA

C

9

Step 5: B’ = 180° - B

B’ = 134.77°

Step 6: C’ = 180° - (A + B’)

B’ = 180° - 45.23° C’ = 180° - (40° + 134.77°)

C’ = 180° - 174.77°C’ = 5.23°

A B’

C’

c’40°

9b = 10

2ND Solution

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Let A = 40°, b = 10, and a = 9

Example 3: SSA

Step 7: a

sin A=

c’

sin C’

9

sin 40°=

c’

sin 5.23°

c’ (sin 40°) = 9 (sin5.23°)

(sin 40°)

c’ = 1.28

(sin 40°)A B’

C’

c’40°

9b = 10

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Let B = 53°, b = 10, and c = 32

Example 4: SSA

Step 1: Solve for h

h = c sin B

h = 32 sin 53°

h = 40.07

b < h ( No Solution) A

C

B

b

a

c

h

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Example 5: SSALet C = 100°, a = 25, and c = 33

Step 1:c

sin C=

a

sin A

33

sin 100°=

25

sin A

33(sin A ) = 25 (sin 100°)

33 33

sin A = 0.75

A = sin-1 0.75

A = 48.59°

Step 2: B = 180° - (A + C)

B = 180° - (48.59° + 100°)

B = 180° - 148.59°

B = 31.41°

C A

B

100°

2533

b

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Example 5: SSALet C = 100°, a = 25, and c = 33

C A

B

100°

2533

Step 3:c

sin C=

b

sin B

33

sin 100°=

b

sin 31.41°

33(sin 31.41° ) = b(sin 100°) sin 100°sin 100°

b = 17.46

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Example 6: SSA

Let A = 133°, a = 27, and c = 40

A

B

C133°

27

40

a < c (No Solution)

Florben G. Mendoza

We use the Law of Sines to solve CASE 1 (SAA or

ASA) and CASE 2 (SSA) of an oblique triangle. The

Law of Cosines is used to solve CASES 3 and 4.

CASE 3: Two sides and the included

angle are known (SAS).CASE 4: Three sides are known (SSS).

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32

Deriving the Law of Cosines

• Write an equationusing Pythagorean theorem for shaded triangle.

b h a

k c - kA B

C

c

sin

cos

h b A

k b A

2 22

2 2 2 2 2 2

2 2 2 2 2

2 2 2

sin cos

sin 2 cos cos

sin cos 2 cos

2 cos

a b A c b A

a b A c c b A b A

a b A A c c b A

a b c c b A

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Law of Cosines

• Similarly

• Note the pattern

2 2 2

2 2 2

2 2 2

2 cos

2 cos

2 cos

a b c c b A

b a c a c B

c b a a b C

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Law of Cosines

a2 = b2 + c2 – 2bc cos A

a2 = b2 + c2 – 2bc cos A

2bc cos A = b2 + c2 – a2 2bc 2bc

cos A = b2 + c2 - a2

2bc

Similarly;

cos A = b2 + c2 - a2

2bc

cos B = a2 + c2 - b2

2ac

cos C = a2 + b2 - c2

2ab

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Example 7: SASLet A = 42°, b = 12.9 & c = 15.4

Step 1: a2 = b2 + c2 – 2bc cos A

a2 = (12.9)2 + (15.4)2 – 2 (12.9) (15.4) (cos 42°)

a2 = 403.57 – 295.27

a2 = 108.3

a =10.41

A B

C

42°

15.4

12.9 a

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Step 2: cos B = a2 + c2 - b2

2ac

cos B = (10.41)2 + (15.4)2 – (12.9)2

2(10.41)(15.4)

Example 3:

Let A = 42°, b = 12.9 & c = 15.4

A B

C

42°

15.4

12.9 a

cos B = 179.12

320.68

cos B = 0.56

B = cos-1 0.56

B = 55.94°

SAS

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Example 3:

Let A = 42°, b = 12.9 & c = 15.4

A B

C

42°

15.4

12.9 a

SAS

Step 3: C = 180° - (A + B)

C = 180° - (42° + 55.94°)

C = 180° - 97.94°

C = 82.06°

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Example 8: SSSLet a = 9.47, b = 15.9 & c = 21.1

Step 1: cos A = b2 + c2 - a2

2bc

cos A = (15.9)2 + (21.1)2 – (9.47)2

2(15.9)(21.1)

cos A = 608.34

670.98

cos A = 0.91

A = cos-1 0.91

A = 24.49°

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Step 2: cos B = a2 + c2 - b2

2ac

Example 8: SSSLet a = 9.47, b = 15.9 & c = 21.1

cos B = (9.47)2 + (21.1)2 – (15.9)2

2(9.47)(21.1)

cos B = 282.08

399.63

cos B = 0.71

B = cos-1 0.71

B = 44.77°

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Example 8: SSSLet a = 9.47, b = 15.9 & c = 21.1

Step 3: C = 180° - (A + B)

C = 180° - (24.49° + 44.77°)

C = 180° - 69.26°

C = 110.74°

C A

B

9.47

21.1

15.9

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