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Section 7.1 & 7.2- Oblique Triangles (non-right triangle)LAW OF SINESIf ABC is a triangle with sides a, b, and c, then
C
c
B
b
A
a
sinsinsin
A B
C
ab
c
h
A B
C
ab
c
h
A, B, and C are acute A is obtuse
c
C
b
B
a
A sinsinsin Two angles and any side: AAS or ASA
Two sides and an excluded angle: SSA (ASS)
A
B
Find the remaining angle and sides for a triangle with the given information.
28
29
102
b
B
C
C102 29
49
a
c
1. Find all angles first. 180 102 29 49A 2. Find sides by using the proportion formula for Law of Sines, use given values!
sin sin sinA B C
a b c
sin(49o) sin(29o) sin(102o)
28
A
a
c
49
29sin49sin28a 6.43
6.43
29sin102sin28c 5.56
5.56
28
Ambiguous Case SSA (ASS)A is an acute angle.
B
C
c
h
Condition a < h a = h a > b
# of Triangles None One One
A
b a
B
C
c
hA
b aB
C
c
hA
b a
Abh sin
Ambiguous Case SSA (ASS)A is an acute angle.
Condition h < a < b
# of Triangles TWO
B
C
c
hA
ba
B’
C’
c’
B
C
c
hA
ba
B’
C’
c’
a
a
A
b
B + B’ = 180o
Ambiguous Case SSA (ASS)A is an obtuse angle.
B
C
c
Condition a < b a > b
# of Triangles NONE One
A
ba
B
C
cA
b a
Find the remaining angle and sides for a triangle with the given information.
25
22
72
b
a
A
Abh sin
B
C
c
hA
b a
Draw the triangle with the given angle in the lower left corner and solve for h.
72sin25 78.23
72
222578.2322
ha
NO TRIANGLE
Find the remaining angle and sides for a triangle with the given information.
12
31
20
c
b
C
B
A
a
hC
b c
20
1231
Draw the triangle with the given angle in the lower left corner and solve for h.
Cbh sin 20sin31 6.10
6.1012 hc
FALSEThere is a triangle.
3112 bc
FALSEMore than 1
triangle.
B
A
a
hC
bc
B’
A’
a’
c’20
3112
Condition h < c < b 10.6< 12 < 31# of Triangles TWO
6.1012
Find the remaining angle and sides for a triangle with the given information.
12
31
20
c
b
C
B
A
c
hC
bc
B’
A’
c’
c20
3112
6.1012
B
A
a
C
b’
c
B’
A’
a’
cC
bB’ = 180o – B
20
20
31
31
12
12
Since the inverse of sine will return an acute angle, we will solve the Acute Triangle first! Find angle B first and then angle A.
b
B
c
C
a
A sinsinsin
sin(20o)
12 31
1220sin31sin B 1.62 1220sin31sin 1 B
1.62
1.6220180
9.97
a
A
B
a
A
B
1.629.97
sin(97.9o)
20sin9.97sin12a 8.34
8.34
9.117
9.117
1.421.42
1.42209.117180
1.62
20sin1.42sin12a 5.23
5.23
b
B
c
C
a
A
sinsinsin
12
sin(42.1o) sin(20o)
Area of an Oblique triangle SAS
B
C
ah
cA
bAbh sin
heightbaseArea2
1
AbcArea sin2
1 BacArea sin
2
1 CabArea sin
2
1
Area of any triangle is one-half the product of the lengths of two sides times the sine of the included angle.
Find the area for a triangle with the given information.
52
90
102
b
a
C
AbcArea sin2
1
CabArea sin2
1 102sin5290
2
1sq87.2288 units
Section 7.3- Oblique Triangles – Law of CosinesSAS & SSS
0,cB
yxC ,
ay
cA
b
b
xA cos
22 0 ycxa
x
Find the distance of “a.”
Substitute the trig. expressions for x and y.
b
yA sin
2222 sincos2cos AbccAbAba
222 0 ycxa2222 2 ycxcxa
Abx cos
Aby sin
Find a trig. expression for x and y.
AbccAbAba cos2sincos 222222 AbcAbcAba 222222 sincos2cos
Abccba
AbccAAba
cos2
cos2sincos222
22222
Standard Form Alternative Form
Cabbac
Baccab
Abccba
cos2
cos2
cos2
222
222
222
ab
cbaC
ac
bcaB
bc
acbA
2cos
2cos
2cos
222
222
222
Find the remaining angles and side of the triangle. SAS
A B
C
b = 9 a
c = 12
25o
1. Find the side opposite the given angle.
Abccba cos2222 25cos1292129 222a
237518.292 a41.5a
41.5
2. Find the angle opposite the shortest given side by the Law of Sines and then subtract the two acute angles from 180o.
c
C
a
A
b
B sinsinsin
sin(25o)
9 5.41 12
41.525sin9sin B
7.44 41.525sin9sin 1 B
7.44
3.110257.44180C
3.110
Find the angles of the triangle. SSS
A
B
C
c = 8 a = 14
b = 19
1. Use the Law of Cosines to find the angle opposite the longest side.
ac
bcaB
2cos
222
8142
19814cos
222
B224
101
224
101cos 1B 8.116B
8.116
2. Find either acute angle by the Law of Sines and then subtract the two angles from 180o.
c
C
b
B
a
A sinsinsin
sin(116.8o)
14 19 8
198.116sin14sin A
1.41 198.116sin14sin 1 A
1.228.1161.41180C
1.41 1.22
2 2 2
2 2 2
2 2 2
2 cos
2 cos
cos2
b a c ac B
b a c ac B
b a cB
ac
Negative value, means Quad. 2 for cos-1x, obtuse angle.
Heron’s Area Formula SAS & SSS
Find the area of the triangle. A B
Cb = 53 a = 43
c = 72
Given any triangle with sides a, b, and c, the area of the triangle is…
where s = ( a + b + c )/2.
csbsassArea
84 84 43 84 53 84 72Area
43 53 72
2s
1. Find the value of s.168
842
84 41 31 12 12 88977 121131.89 .sq units
Section 7.4 - Vectors in the PlaneForce and velocity involve both magnitude (distance) and direction (slope) and cannot be completely characterized by a single real number. We will use a DIRECTIONAL LINE SEGMENT (RAY) to represent force and velocity (vectors).
P
Initial Point
Q
Terminal PointPQ ��������������
v
Let u represent the directed line segment from P(0,0) to Q(3,2) and v be the directed line segment from R(1,2) to S(4,4). Show they are equivalent.
P
QR
S
u
v
130203 22 u
132414 22 v
Same Magnitude
3
2
03
02
um
3
2
14
24
vm
Same Direction
Equivalent vectors must have the same magnitude and direction.
The multiplication of a real number k and a vector v is called scalar multiplication. We write this product kv.
Multiplying a vectorby any positive realnumber (except 1)changes the magnitudeof the vector but not its direction.
Multiplying a vector by any negativenumber reversesthe direction of the vector.
Parallelogram LawConnecting the terminal point of the first vector with the initial point of the second vector to obtain the sum of two vectors.
u
v
u + v
Find u - v
The sum of u and v, denoted u + v is called the resultant vector. A geometric method for adding two vectors is shown in the figure.
Here is how we find this vector:
u
v- v
u – v
The i and j Unit Vectors
The Vector i is the unit vector whose direction is along the positive x-axis.
Vector j is the unit vector whose direction is along the positive y-axis.
1
1
j
ix
y
Consider the point P(a, b). The initial point is at (0, 0) and terminal point is
at (a, b). Vector v can be represented as v = ai + bj.
x
y ),( baP
ai
bjv = ai + bj
Another notation is position
vector form. v ba,
u = (-2 – 3)i + (5 – (-1))j
u = – 5i + 6j u 5,6
Adding and Subtracting Vectors in Terms of i and j
If v = 7i + 3j and w = 4i – 5j, find the following vectors:
a. v + w
b. v – w
= (7i + 3j) + (4i – 5j)
Comb. Like Terms = 11i – 2j
= (7i + 3j) – (4i – 5j)
= 3i + 8j = 7i + 3j – 4i + 5j
Dist. Prop of minus sign
Position vector form.
v = w = 11,ba 22 ,ba
v + w = 2121 , bbaa v – w = 2121 , bbaa
v = w = 3,7 5,4
v + w = 53,47
2,11
v – w = 53,47
8,3
Scalar Multiplication with a Vector in Terms of i and j
If v = 7i + 3j and w = 4i – 5j, find the following vectors:
a. 3v
b. 3v – 2w
= 3(7i + 3j)
= 21i + 9j
= 3(7i + 3j) – 2(4i – 5j)
= 21i + 9j – 8i + 10j
= 13i + 19j
v = w = 3,7 5,4
3v = 3,73 33,73 9,21
3v – 2w = 5,423,73 21,9 8, 10
21 8,9 10 13,19
21,9 8,10
21 8,9 10 13,19
Or distribute the minus sign
0,0
Find the unit vector in the same direction as v = 4i – 3j. Then verify that the vector has magnitude 1.
2 2v a b 2 24 ( 3) 16 9 25 5
4 35
v i jv
2 24 35 5
4 35 5
i j
16 925 25
251
25
3,4
53
54 ,
Writing a Vector in Terms of Its Magnitude and Direction
v
acos
v
bsin
v = ai + bj
cosva sinvb
v = i + jcosv sinv
sin,cos vv
Remember your identities…sin
tancos
bv ba av
v = ai + bj
a
b
The jet stream is blowing at 60 miles per hour in the direction N45°E. Express its velocity as a vector v in terms of i and j.
x
y
v
45
N
45
E
45 , 60v
cos sinv v i v j
60cos45 60sin 45v i j 2 2
60 602 2
i j
30 2 30 2i j
The jet stream can be expressed in terms of i and j as
v = i + j230 230
The dot product of two vectors results in a scalar (real number) value, rather than a vector.
If v = 7i – 4j and w = 2i – j, find each of the following dot products:
a.
b.
c.
v w 1427 18414
w v 4172 18414
w w 1122 514
Alternative Formula for the Dot Product
Find the angle between the two vectors v = 4i – 3j and w = i + 2j. Round to the nearest tenth of a degree.
wv
wv1cos
2222
1
2134
2314cos
125
2cos
525
2cos 11
100.3
The angle betweenthe vectors is
100.3 .
Two forces, F1 and F2, of magnitude 30 and 60 pounds, respectively, act on an object. The direction of F1 is N10°E and the direction of F2 is N60°E. Find the magnitude, to the nearest hundredth of a pound, and the direction angle, to the nearest tenth of a degree, of the resultant force.
10N E
60N E301 F
602 F
1 1 1 1 1cos sinF F i F j
30cos80 30sin80i j
5.21 29.54i j
2 2 2 2 2cos sinF F i F j
60cos30 60sin30i j
51.96 30i j
F
x
y
1 80 2 30
1 5.21 29.54F i j 2 51.96 30F i j
1 2F F F (5.21 51.96) (29.54 30)i j
57.17 59.54i j
2 2F a b 2 257.17 59.54 82.54
cosaF
57.1782.54
1 57.17cos
82.54
2.46
43.8N E
90 46.2 43.8
10N E
60N E301 F
602 F
F
x
y
v
acos
v
bsin
tanb
a
Identities
Parallel and Orthogonal VectorsTwo vectors are parallel when the angle between the vectors is 0° or 180°. If = 0°, the vectors point in the same direction. If = 180°, the vectors point in opposite directions.
1
wv
wv
1
wv
wv
Two vectors are orthogonal when the angle between the vectors is 90°.
0 wv
Are the vectors v = 2i + 3j and w = 6i – 4j orthogonal?
v w 012124362 Yes
Dot Product of Acute Angles are positive,
Right Angles are zero, and
Obtuse Angles are negative.
Using Vectors to Determine Weight.A force of 600 pounds is required to pull a boat and trailer up a ramp inclined 15o from the horizontal. Find the combined weight of the boat and trailer.
WB
CA
AC
BC
BA Force of gravity = combined weight.
Force against ramp.
Force required to move boat up ramp = 600lbs
600sin 15
6002318
sin 15
AC
BA BA
BA lbs
15
15
600
Using Vectors.A plane is flying at a bearing N 30o W at 500 mph. At a certain point the plane encounters a 70 mph wind with the bearing N 45o E. What are the resultant speed and direction of the plane?
120o
u
v
u + v
45o
500 cos120 ,sin120
1 3500 , 250,250 3
2 2
u
70 cos 45 ,sin 45
2 270 , 35 2,35 2
2 2
v
2 2
250 35 2,250 3 35 2 200.5,482.5
200.5 482.5 522.5
u + v
1
482.5tan
200.5482.5
tan 67.4200.5
b
a
180 67.4 112.6
22.6N W
||u+v||
Using Vectors to find tension.Determine the weight of the box.
??? lbs
50o 30o
A
C
B
879.4 lbs652.7 lbs
310 210
879.4cos 310 879.4sin 310i j u
652.7cos 210 652.7sin 210i j v
u + v 879.4cos 310 652.7cos 210 i 879.4sin 310 652.7sin 210 j
u + v 2 2A B
1000 lbs
