module 8 lesson 5 oblique triangles florben g. mendoza

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

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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)

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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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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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