7. Triangles
7.1
7.2
7.3
7.4
7.5
7.6
The Law of Sines
The Ambiguous Case
The Law of Cosines
The Area of a Triangle
Vectors: An Algebraic Approach
Vectors: The Dot Product
1
7. Triangles
Solving Triangles
Case
1. AAA
2. AAS
3. ASA
4. SSA
5. SAS
6. SSS
Solvability
no
yes
yes
ambiguous case*
yes
yes
Method
none
law of sines
law of sines
law of sines
law of cosine
law of cosine
* In this case, there are three possible solutions.
2
7.1 The Law of Sines
1)
2)
3)
What is the Law of Sines
The determined case: AAS or ASA
Problems (method: law of sines)
3
7.1 The Law of Sines
1)
What is the Law of Sines
The Law of Sines
sin A sin B sin C
=
=
a
b
c
C
or, equivalently,
a
b
c
=
=
sin A sin B sin C
ƒ
ƒ
a
b
A
c
B
Easy to prove
Easy to remember
4
7.1 The Law of Sines
2)
The determined case: AAS or ASA
•
Once two angles and one side is know, the remaining
angle and two sides are uniquely determined.
Note that once two angles are known, the third angle
is determined
This include case 2 and case 3.
•
•
5
7.1 The Law of Sines
3) Problems (using significant digit rule, section 2.3)
(1) In ΔABC, A = 33°, C = 82°, and b = 18 cm; find B
and then find c.
[10]
B = 65°, c ≈ 20 cm
(2) In ΔABC, A = 105°, B = 45°, and c = 630 cm; find all
missing parts.
[18]
C = 30°, a ≈ 1200 cm, b ≈ 890 cm
6
7.1 The Law of Sines
3) Problems (using significant digit rule, section 2.3)
(3) Refer to the following figure.
[24]
C
r
r s
D
B
x
A
y
A = 55°, s = 21, and r = 22, find y.
s
21
C = =
≈ 55 o
r
22
y ≈ 22
7
7.1 The Law of Sines
3) Problems (using significant digit rule, section 2.3)
(4) A person standing on the street looks up to the top of a
building and find the angle of elevation is 38°. She
then walks one block further away (440 ft) and finds
that the angle of elevation to the top of the building is
now 28°. How far away from the building is she
makes her second observation? [26]
θ = 142°; α = 10°;
α
c
c = 440sin(142°)/sin(10°)
38°
28° θ
≈ 1560 ft
440 ft
Distance = 1560cos(28°) ≈ 1400 ft
8
7.2 The Ambiguous Case
1)
2)
The ambiguous case: SSA
Problems (method: law of sines)
•
Start with drawing a triangle, label sides and angles, and
indicate the given values
9
7.2 The Ambiguous Case
1) The ambiguous case: SSA
Three possibilities:
i.
ii.
iii.
No solution
One solution
Two solutions
2
6
A
30°
i. No solution
6
A
6
a=h
30°
B
ii. One solution
A
a
h
a
30°
B’
B
iii. Two solutions
10
7.2 The Ambiguous Case
2) Problems
Solve the triangle for each set of values.
(1) A = 150°, b = 30 ft, and a = 10 ft
[2]
Ans. no solution.
start with solving B.
(2) A = 30°, b = 12 cm, and a = 6 cm
[4]
start with solving B. Ans. one solution.
B = 90 o , C = 60 o , c = 6 3 cm
11
7.2 The Ambiguous Case
2) Problems
Solve the triangle for each set of values.
(3) A = 43°, a = 31 ft, and b = 37 ft
[8]
two solution:
Solution I: B ≈ 54°, C ≈ 83°, c ≈ 45 ft
Solution II: B’ ≈ 126°, C ≈ 11°, c ≈ 8.7 ft
(4) B = 62°40′, b = 6.78 in, and c = 3.48 in
[14]
(always check if sum of three angles = 180°)
B = 62.7°. Solution I: C ≈ 27.1°, A ≈ 90.2°, a ≈ 7.64 in
Solution II: C’ ≈ 152.9°, not valid
12
7.2 The Ambiguous Case
3) Navigation
Definition. The heading of an object is the angle,
measured clockwise from due North, to the vector
representing the intended path of the object.
Definition. The true course of an object is the angle,
measured clockwise from due North, to the vector
representing the actual path of the object.
13
7.2 The Ambiguous Case
3) Navigation (more terminology)
With the specific case of objects in flight
–
–
–
–
Airspeed: the speed of the object relative to the air.
Airspeed is the magnitude of the vector representing the
velocity of the object in the direction of the heading.
Ground speed: the speed of the object relative to the ground.
Ground speed is the magnitude of the vector representing the
velocity of the object in the direction of the true course.
14
7.2 The Ambiguous Case
3) Navigation (more terminology)
Ex. True Course. A plane headed due east is traveling with
an airspeed of 190 mph. The wind currents are moving
with constant speed in the direction of 240°. If the
ground speed of the plane is 95 mph, what is the true
course?
[30]
first find α, using law of sines.
then find β,
190 mph
wind
240°
β
30°
α
True course
95 mph
α = 90°,
β = 60°
True course = 150°
15
7.2 The Ambiguous Case
3) Navigation (more terminology)
Ex. Current. A ship is headed due north at a constant 16
miles per hr. Because of the ocean current, the true
courses of the ship is 15°. If the currents are a constant
14 miles per hour, in what direction are the currents
running?
[28]
first find α, using law of sines.
β
14
then find β, which is current’s direction
α
16
15°
α = 17° or 163 °,
β = 32° or 178 ° from due north
16
7.3 The Law of Cosines
1)
2)
3)
What is the Law of Cosines
The determined case: SAS or SSS
Application
17
7.3 The Law of Cosines
1) What is the Law of Cosines
The Law of Cosines
C
a2 = b2 + c2 – 2bc cos(A)
a
b
b2 = a2 + c2 – 2ac cos(B)
A
c
B
c2 = a2 + b2 – 2ab cos(C)
ƒ
ƒ
Easy to prove
Easy to remember
18
7.3 The Law of Cosines
2) The determined case: SAS or SSS
(SSS)
(1) Ex. If a = 51 cm, b = 24 cm, and c = 31 cm, find the
largest angle.
[8]
Ans. 136°
(2) Ex. Solve ΔABC if a = 76.3 m, c = 42.8 m, B = 16.3°.
[10]
Ans. b ≈ 37.2 m, A ≈ 144.9°, C ≈ 18.8°
19
7.3 The Law of Cosines
3) Application
(3) Ex. Distance between two ships. Two ships leave a
harbor entrance at the same time. The first ship is
traveling at a constant speed 18 miles per hour, while
the second is traveling at a constant 22 miles per
hour. If the angle between their courses is 123°, how
far apart are they after 2 hours.
[22]
Ans. ≈ 70 mi
20
7.3 The Law of Cosines
3) Application
(4) Ex. True course and speed. A plane is flying with
airspeed of 244 mph with heading 272.7°. The wind
currents are running at a constant 45.7 mph in the
direction 262.6°. Find the ground speed and the true
courses of the plane.
[26]
N
C = 180° – (2.7° + 7.4°) = 169.9°
45.7 C
244
c
θ
45.7
Ans. ground speed = |c| ≈ 189 mph
θ ≈ 1.72°
Ans. true course ≈ 272.7 – 1.72 ≈ 271°
21
7.4 The Area of a Triangle
1)
2)
3)
Area formula for SAS
Area formula for ASA
Area formula for SSS
22
7.4 The Area of a Triangle
1) Area formula for SAS
Case I: two sides and the included angle (SAS)
1
S = bc sin A
2
1
S = ab sin C
2
1
S = ac sin B
2
ƒ
ƒ
C
a
b
A
c
B
Easy to prove (area = half of base time height)
Easy to remember (?)
23
7.4 The Area of a Triangle
2) Area formula for ASA
Case II: two angles and the included side (ASA)
a 2 sin B sin C
S=
2 sin A
b 2 sin A sin C
S=
2 sin B
2
c sin A sin B
S=
2 sin C
ƒ
ƒ
C
a
b
A
c
B
Easy to prove (using case I and law of sines)
Easy to remember (?)
24
7.4 The Area of a Triangle
3) Area formula for SSS
Case III: three sides (SSS)
C
S = s ( s − a )( s − b)( s − c)
where,
a
b
a+b+c
s =
2
A
c
B
i.e. s is the half of the perimeter.
ƒ
ƒ
Easy to prove (combining I and II, & algebra)
Easy to remember (?)
25
7.4 The Area of a Triangle
4) Problems.
Find the area of each ΔABC.
(1) a = 76.3 m, c = 42.8 m, and B = 16.3°. (SAS)
[4]
Ans. 458 m2
(2) B = 57°, C = 31° and a = 7.3 m. (ASA) [8]
Ans. 11.5 m2
(3) a = 48 yd, b = 75 yd, and c = 63 yd. (SSS)
s = 458 m
[16]
Ans. 1,500 m2
26
7.5 Vectors: An Algebraic Approach
1)
2)
3)
Review: a geometric approach to vectors
An algebraic approach to vectors
Problems
27
7.5 Vectors: An Algebraic Approach
1) Review: a geometric approach to vectors
• Two ingredients in a vector
1) Magnitude
2) Direction
•
•
•
Addition and subtraction of vectors
Horizontal and vertical components of a vector
Standard position – recall that a vector v is in standard
position if the tail of the vector is at the origin.
y
vy
θ
v
vx
x
28
7.5 Vectors: An Algebraic Approach
2) An algebraic approach to vectors
• Vector in standard position
y
(a, b)
vy
θ
v
vx
•
•
•
x
It is a ray from origin (0, 0) to a point (a, b)
Where |vx| = a, |vy| = b.
We write the vector in coordinate form:
–
–
v = 〈a, b〉 or v = ai +bj
i is the unit vector from (0, 0) to (1, 0), and j is the unit
vector from (0, 0) to (0, 1).
29
7.5 Vectors: An Algebraic Approach
2) An algebraic approach to vectors
Magnitude |v| = a 2 + b 2
−1
Direction θ = tan ba
Add and subtract vectors
•
•
•
–
•
y
vy= bj
θ
v
vx= ai
Add or subtract components
x
y
Multiply a vector by a scalar
–
Multiply each component
j
i
x
30
7.5 Vectors: An Algebraic Approach
3) Problems
(1) Draw a vector v from (0, 0) to (5, –5).
[6]
(2) Draw a vector v from (0, 0) to (–5, –1). Then write v in
terms of the unit vectors i and j. [16]
(3) Find the magnitude of the given vector.
(a) 〈–3, 7〉 [18]
(b) 3i + j
Ans. 58
[28]
Ans. 10
31
7.5 Vectors: An Algebraic Approach
3) Problems
(4) Let u = 〈1, 4〉 and v = 〈–2, 5〉 . Find
(a) u + v
(b) u – v
〈–1, 9〉
〈3, –1〉
(5) Let u = 5i + 3j, v = 3i + 5j. Find
(a) u + v
(b) u – v
8i + 8j
2i – 2j
[34]
(c) 2u – 3v
〈8, –7〉
[40]
(c) 3u + 2v
21i + 19j
32
7.5 Vectors: An Algebraic Approach
3) Problems
(6) vector u is in standard position, and makes an angle of
110° with the positive x-axis. Its magnitude is 25.
Write u in component form 〈a, b〉 and vector form
ai + bj.
[42]
Component form: 〈–8.6, 23〉; vector form –8.6i + 23j
(7) Let v = –2 3 i + 2j. Find the magnitude of v and the
angle θ, 0° ≤ θ < 360°, that the vector v makes with
the positive x-axis.
[48]
Magnitude = 4, θ = 150°
33
7.6 Vectors: The Dot Product
1)
2)
3)
4)
5)
What is dot product?
Finding the angle between two vectors
Perpendicular vectors
Application: work
problems
34
7.6 Vectors: The Dot Product
1) What is dot product?
Definition
The dot product of two vectors u = ai + bj and
v = ci + dj is written u•v and is defined as:
u•v = (ai + bj)•(v = ci + dj)
= ac + bd
Note, the dot product is a real number, not vector
35
7.6 Vectors: The Dot Product
2) Finding the angle between two vectors
Theorem 7.1
The dot product of two vectors u and v is equal to the
product of their magnitudes multiplied by the cosine
of the angle between them, i.e.
u•v = |u||v|cosθ
where θ is the angle between u and v.
We can also solve the angle between u and v by
u •v )
θ = cos–1( |u||v|
36
7.6 Vectors: The Dot Product
3) Perpendicular vectors
Theorem 7.2
If u and v are two nonzero vectors, then
u•v = 0 ⇔ u ⊥ v.
In words, two nonzero vectors u and v are
perpendicular if and only if their dot product is zero.
37
7.6 Vectors: The Dot Product
4) Application: Work.
Work is performed when a constant force F (in pounds) is
used to move an object a certain distance (in feet).
F
d
Theorem 7.3
If a constant force F is applied to an object, and the
resulting movement of the object is represented by the
displacement vector d, then the work performed by the
force is
Work = F•d (in ft-lb)
38
7.6 Vectors: The Dot Product
5) Problems
(1) Find the dot product 〈3, 4〉•〈5, 5〉
[2] Ans. 35
(2) Find u•v if u = –i + j and v = i + j
[6]
Ans. 0
(3) Find the angle θ between u = –3i + 5j and v = 6i + 3j.
[14]
Ans. ≈ 95°
(4) Find the value of a such that u = ai + 6j and
v = 9i + 12j are perpendicular.
Ans. a = –8
39
7.6 Vectors: The Dot Product
5) Problems
(4) A force F = 45i – 12j (lb) is applied to an object,
displacing it according to the vector d = 70i + 15j
(feet). Find the amount of work done.
Ans. work = 2,970 ft-lb
(5) A package is pushed across a floor a distance of 52ft by
a force of 15 lb exerted downward at an angle of 25°
from the horizontal. Find the work done.
Ans. work = 710 ft-lb
40