Module 18 Oblique Triangles (Applications) Florben G. Mendoza.
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Transcript of Module 18 Oblique Triangles (Applications) Florben G. Mendoza.
Module 18Oblique Triangles
(Applications)
Florben G. Mendoza
FOUR CASES
CASE 1: One side and two angles are known (SAA or
ASA). Law of Sines
CASE 2: Two sides and the angle opposite one of them
are known (SSA). Law of Sines
CASE 3: Two sides and the included angle are known
(SAS). Law of Cosines
CASE 4: Three sides are known (SSS). Law of Cosines
Florben G. Mendoza
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Law of Sines Law of Cosines
a
sin A=
b
sin B
c
sin B=
a2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C
cos A = b2 + c2 - a2
2bc
cos B = a2 + c2 - b2
2ac
cos C = a2 + b2 - c2
2ab
Florben G. Mendoza
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Practical applications of trigonometry often involve
determining distances that cannot be measured directly. In
many applications of trigonometry the essential problem is the
solution of triangles.
If enough sides and angles are known, the remaining sides and
angles as well as the area can be calculated, and the triangle is
then said to be solved. Problems
involving angles and distances in one plane are covered in this
lesson.
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5
Example 1: A navy aircraft is flying over a straight highway. When
the aircraft is in between the two cities that are 5 miles apart, he
determines that the angle of depression to two cities to be 32° and
48° respectively. Find the distance of the aircraft from the two
cities.
32° 48°
5 mi
32° 48°
A B
C
a = ?b = ? City BCity A
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32° 48°
5 mi
a = ?b = ?
A B
C
A = 32° B = 48° c = 5 mi
Given:
Find:a = ? b = ?
Step 1: ASA – Law of Sine
Step 2: C = 180° - (32° + 48°)
C = 180° - (A+B)
C = 100°
C = 180° - 80°
Step 3: c
sin C=
a
sin A
5
sin 100°=
a
sin 32°
a (sin 100°) = 5 (sin 32°)
sin 100° sin 100°
a = 2.69 mi
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32° 48°
5 mi
a = ?b = ?
A B
C
A = 32° B = 32° c = 5 mi
Given:
Find:a = ? b = ?
Step 4: c
sin C=
b
sin B
5
sin 100°=
b
sin 48°
b (sin 100°) = 5 (sin 48°)
sin 100° sin 100°b = 3.77 mi
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Example 2: Three circles of radii 100, 140, & 210 cm
respectively are tangent to each other externally. Find the
angles of the triangle formed by joining their centers.
A
C
B240
35031021
0 210
C
140
140
B
100
100
A ?
?
?
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A
C
B240
350310
Step 2: cos A = b2 + c2 - a2
2bc
cos A = (310)2 + (240)2 – (350)2
2(310)(240)
cos A = 31 200
148 800
cos A = 0.21
A = cos-1 0.21
A = 77.88°
Given:
a = 350
b = 310
c = 240
Find:
A = ?
B = ?
C = ?
Step 1: SSS – Law of Cosine
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Step 3: cos B = a2 + c2 - b2
2ac
cos B = (350)2 + (240)2 – (310)2
2(350)(240)
cos B = 161 000
217 000
cos B = 0.74
B = cos-1 0.74
B = 42.27°
A
C
B240
350310
Step 4:
C = 180° - (77.88° + 42.27°)
C = 180° - (A+B)
C = 59.85°
C = 180° - 120.15°
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Example 3: A pole casts a shadow of 15 meters long when the angle
of elevation of the sun is 61°. If the pole has leaned 15° from the
vertical directly towards the sun, find the length of the pole.
15 m
61°
15° ?
15 m
61° 105°
A
B
C
?
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105°61°
15 m
?
B
A C
Given:
A = 61°
C = 105°
b = 15 m
Find:
a = ?
Step 1: ASA – Law of Sine
Step 2: B = 180° - (61° + 105°)
B = 180° - (A + C)
B = 14°
B = 180° - 166°
Step 3: b
sin B=
a
sin A
15
sin 14°=
a
sin 61°
a (sin 14°) = 15 (sin 61°)
sin 14° sin 14°
a = 54.23 m
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Example 4: A tree on a hillside casts a shadow 215 ft down the
hill. If the angle of inclination of the hillside is 22° to the
horizontal and the angle of elevation of the sun is 52°, find the
height of the tree.
52 22 215 ft
22
30 38
22
68
112
38
215 ft
30
112
38
215 ft
A
C
B
?
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Given:
A = 30°
B = 112°
C = 38°
Find:
a = ?
A
B
C
30°
38°
112°
215 ft
?c
sin C=
a
sin A
215
sin 38°=
a
sin 30°
a (sin 38°) = 215 (sin 30°)
sin 38° sin 38°
a = 174.61 ft
c = 215 ft
Step 1:
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Example 5: A pilot sets out from an airport and heads in the
direction N 20° E, flying at 200 mi/h. After one hour, he makes a
course correction and heads in the direction N 50° E. Half an
hour after that, engine trouble forces him to make an emergency
landing. Find the distance between the airport & his final landing
point.
20°
50°
200
mi
100 mi
?
20°
40°
200
mi
100 mi
150°
?
A
B C
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200
mi
100 mi
150°
?
A
B C
Given:
B = 150°
a = 100 mi
c = 200 mi
Find:
b = ?
Step 2:
SAS – Law of CosineStep 1:
b2 = a2 + c2 – 2ac cos B
b2 = (100)2 + (200)2 – 2(100)(200) (cos 150°)
b2 = 50 000 – (- 34 641.02)
b2 = 84 641.02
b = 290.93
Florben G. Mendoza