module 8 lesson 5 oblique triangles florben g. mendoza
TRANSCRIPT
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Module 8Lesson 5
Oblique Triangles
Florben G. Mendoza
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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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