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C. C. a. b. a. b. B. B. A. c. c. A. 7.1 & 7.2 Law of Sines. Oblique triangle – A triangle that does not contain a right angle. sin A sin B sin C a b c. =. =. or. a b c__ sin A sin B sin C. =. =. C. b. a. B. A. 14cm. - PowerPoint PPT Presentation
7.1 & 7.2 Law of SinesOblique triangle – A triangle that does not contain a right angle.
C
BA
b a
c A
B
C
c
a
b
sin A sin B sin C a b c
= =
or
a b c__sin A sin B sin C
= =
Solving SAA or ASA TrianglesSAA and ASA Triangles – Oblique triangles in which you know 2 angles and 1 side.
C
BA
b a
14cm
sin A sin B sin C a b c
= = or a b c__sin A sin B sin C= =
64°
82°
Solving for <B There are 180 degrees in a triangle, so 64 + 82 + B = 180 146 + B = 180 B = 34°
Solving for a Solving for b Sin (82) = Sin (64) Sin(82) = sin(34) 14 a 14 b
aSin(82) = 14 Sin(64)
a = 14 Sin(64) Sin (82) a = 12.7cm b = 7.9 cm
Try this one on your own
A
B
C
c
a
20
sin A sin B sin C a b c
= = or a b c__sin A sin B sin C= =
102° 38°
Check your answers: < C = 40° a = 31.8 c = 20.9
Application Problem: Bearings
• The bearing of a lighthouse from a ship was found to be N 52° W. After the ship sailed 5.8 km due south, the new bearing was N 23° W.
• Draw a Picture to represent the situation
• Find the distance between the ship and the lighthouse at each location.
sin A sin B sin C a b c
or a b c__sin A sin B sin C
7-5
• The bearing of a lighthouse from a ship was found to be N 52° W. After the ship sailed 5.8 km due south, the new bearing was N 23° W.
7.1 Example 3 Using the Law of Sines in an Application
(ASA) (page 305)
Let x = the distance to the lighthouse at bearing N 52° W andy = the distance to the lighthouse at bearing N 23° W.
Application Problem: Bearings
• Jerry wishes to measure the distance across the Big Muddy River. He determines that C = 117.2°,
A = 28.8°, and b = 75.6 ft. Find the distance a across the river.
7.1 Example 2 Using the Law of Sines in an Application
(ASA) (page 305)
Practice: Law of Sines
SSA Ambiguous Case TrianglesSSA Triangles – Oblique triangles in which you know 2 of the sides and an angle opposite one of the known sides.
You may have NO solutions, ONE solution, or TWO solutions
Example (NO solutions)
Solve the triangle ABC if A = 75°, a = 51, and b = 71
sin A sin B sin C a b c
SSA Ambiguous Case(1 & 2 Solution Triangles)
Example (ONE solution)
Solve the triangle ABC A = 61°, a = 55, and c = 35
sin A sin B sin C a b c Example (TWO solutions)
Solve the triangle ABC A = 40°, a = 54, and b = 62
Solve triangle ABC if B = 68.7°, b = 25.4 in., and a = 19.6 in.
Practice1:
• Solve triangle ABC if A = 61.4°, a = 35.5 cm, and b = 39.2 cm.
Practice2:
Area of Oblique Triangles
C
BA
b a
c
sin A = h b
h = b SinAh
Since Area = ½ (Base)(Height) Area = ½ c (b SinA) Area = ½ b c Sin A
You may use any one of 3 area formulas.Area = ½ b c Sin AArea = ½ a b Sin CArea = ½ a c Sin B
Example: (Find the Area)A = 62° b = 10 c = 24 m
Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-12
• Find the area of triangle DEF in the figure.
Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-13
• Find the area of triangle ABC if B = 58°10′, a = 32.5 cm, and C = 73°30′.
We must find AC (side b) or AB (side c) in order to find the area of the triangle.
7.3 Law of CosinesC
BA
b a
c
a2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C
A
B
C
c
a
b
Solving SAS TrianglesC
BA
b=20 a
c=30
60°
Law of Cosinesa2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C
Law of Sinessin A sin B sin C a b c= =
Step 1: Use Law of Cosines to find side opposite given angle a2 = b2 + c2 – 2bc cos A a2 = (20)2 + (30)2 – 2(20)(30) cos (60) a = 26 Step 2: Use Law of Sines to find angle opposite the shorter of the 2 given sides.
Step 3: Find the 3rd angle in the triangle using concept of 180 degrees in a triangle.
sin B = sin (60) 20 26
B = 41°
C = 79°
Try this one on your own
A
B
C
8
a
7
Law of Cosinesa2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C
Law of Sinessin A sin B sin C a b c
120°
Check your answers: < B = 28° a = 13 <C = 32°
Solving SSS TrianglesLaw of Cosinesa2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C
Law of Sinessin A sin B sin C a b c= =
Step 1: Use Law of Cosines to find angle opposite the longest side b2 = a2 + c2 – 2ac cos B 2ac cos B = a2 + c2 - b2
cos B = a2 + c2 - b2
2ac cos B = 62 + 42 – 92 = -29/48 B = 127.2 2(6)(4)
Step 2: Use Law of Sines to find either of the other angles. A = 32.1Step 3: Find the 3rd angle in the triangle using concept of 180 degrees in a triangle. C = 20.7
B
A
C
c=4
b=9
a=6
Try this one on your ownLaw of Cosinesa2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C
Law of Sinessin A sin B sin C a b c= =
B
A
C
c=8
b=16
a=10
Heron’s Formula for Triangle Area(Derived from Law of Cosines)
If a triangle has sides of lengths a, b, c with semiperimeter s = (1/2)(a + b + c)
Area = s(s - a)(s - b)(s – c)
Example: Find the area of the triangular region:
A
CB
c=2427 mi b=331 miles
a=2451 miles
S = (1/2)(2451+331+2427) = 2604.5
A = 2604.5 (2604.5 –2451)(2604.5 –331)(2604.5 –2427)
A = 401,700 mi2
7-4 & 7.5 Vectors
Scalar – A quantity that involves magnitude, but no direction. (Example: the temperature outside might be 75°)
Vector – A quantity that involves both magnitude and direction. (Example: The jet is flying 120 mph 30° East of North)
Magnitude – How big something is – the size
Direction – The angle in degrees something is travelling
Vectors can be represented as directed line segments• Magnitude = length of segment (distance formula)• ||v|| or | v| - the magnitude of vector v
vu
Showing 2 Vectors are Equal
Two vectors (u and v) are equal if they have the same magnitudeand the same direction.
• Magnitude Use the distance formula to see if ||u|| = ||v||
• Direction Find the slope of each vector. If the slopes are the same u and v have the same direction.
•Example Let u have initial point (-3, -3) and terminal point (0, 3) Let v have initial point (0, 0) and terminal point (3, 6)
Does u = v? YES! - ||v|| = ||u|| = 45 = 35 and both slopes = 2
Scalar Multiplication (Geometric)
v2v ½ v -2v
Multiplying a vectorby a negative numberreverses the direction
Multiplying a vector by a positivenumber changes the magnitudebut not the direction
If k is a real number and v is a vector, then kv is a scalar multiple of vector v.
A geometric method for adding two vectors is shown below. The sum of u + v is called the resultant vector. Here is how we find this vector.
1. Position u and v so the terminal point of u extends from the initial point of v.
2. The resultant vector, u + v, extends from the initial point of u to the terminal point of v.
Initial point of u
u + vv
u
Resultant vector
Terminal point of v
The Geometric Method for Adding Two Vectors
Opposite vectors: v and –v-- Same magnitude, opposite direction-- The sum of opposite vectors has magnitude 0 – called the zero vector
v -v
1
1
i
j
Ox
y
Vectors in the X/Y Axes• 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.
Position Vector: Vector with initial point at originPosition Vector, v, with endpoint at (a,b) is written <a, b>
i and j are unit vectors (they have magnitude of 1)i = <0, 1> and j = <1, 0>
1
1
i
j
Ox
yVectors are also represented in terms of i and j.
A vector, v, with initial point (0,0) andtermianl point (a,b) is also represented <a, b>
Magnitude:
v a2 b2
a
bj
v=ai+bj
(a, b)
(Pythagorean Theorem)
a = ||u|| cos b = ||u|| sin
v = <a, b> = <||u|| cos , ||u|| sin >
Example 1: Sketch the vector <-3. 4> , v = -3i + 4j and find its magnitude.
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1-2
-3
-4-5
Initial point
Terminal point
v = -3i + 4j
v a2 b2
( 3)2 42
9 16
25 5
Vector Representation & ExamplesGeneral formula for representation of a vector v with initial point P1 = (x1, y1)
and terminal point P2 = (x2, y2) is equal to the position vector
<(x2 – x1) , (y2 – y1) > and v = (x2 – x1)i + (y2 – y1)j.
Example 2: Vector v has initial point (3, -1) and terminal point (-2, 5)
a) Represent it in i, j format b) Represent it in <x , y> form c) Find magnitude ||v||
v = (-2 – 3)i + (5 – (-1))jv = -5i + 6j 61<-5, 6>
Vector Operations: Addition, Subtraction, and Scalar Multiplicationv = a1i + b1j and w = a2i + b2j
v + w = (a1 + a2)i + (b1 + b2)j
v – w = (a1 – a2)i + (b1 – b2)j
Example: v = 5i + 4j and w = 6i – 9j
v + w = __________________ v – w = ___________________
Let k be a constant and v = a1i + b1j then kv = (ka)i + (kb)j
Example: v = 5i + 4j
6v = ___________________ -3v = __________________
11i – 5j -i +13j
30i + 24j-15i - 12j
Vector Operations Revisitedv = <a, b> and w = <c, d>
v + w = <(a + c), (b + d)>
v – w = <(a – c), (b – d)>
Example: v = <5, 4> and w = <6, -9>
v + w = __________________ v – w = ___________________
Let k be a constant and v =< a , b> then kv = <(ka), (kb)>
Example: v = <5, 4>
6v = ___________________ -3v = __________________
<11 – 5> <-1, 13>
<30, 24> <-15, – 12>
If v = <a, b> then –v = <-a, -b> Example: v = <6, -2>, so –v = <-6, 2>
Zero Vector and Unit VectorThe vector whose magnitude is 0 is called the zero vector. The zero vector has no direction: 0 = 0i + 0j = <0, 0>
The vector whose magnitude is 1 is called the unit vector.To find the unit vector in the same direction as a given vector v,divide v by its magnitude
Unit Vector = _v_ ||v||
Example: Find the unit vector in the same direction as v = 5i – 12j
First find ||v|| = || 5i –12j|| =52 + 122 = 25 + 144 = 169 = 13
Unit Vector = _v_ = 5i – 12j = _5 i - 12 j ||v|| 13 13 13
Using Magnitude and Direction to write a Vector
(a,b)
||v|| v = ai + bj
v = ||v||cosi + ||v||sinj
= < ||v||cos, ||v||sin>Example:Wind is blowing at an angle = 120 Magnitude ||v|| = 20mphExpress the wind’s velocity as a vector
v = ||v||cosi + ||v||sinj
v = 20cos(120)i + 20sin(120)j
v = 20(- ½ )i + 20 (3/2)j
v = -10i + 103 j = <-10, 103 >
Dot Product
v = <a, b> w = <c, d>
The dot product v • w is defined as follows:
2121 bbaawv
Examples: v = <5, – 2> w = <-3, 4>
v · w = 5(-3) + (-2)(4) = -15 – 8 = -23
w · v = (-3)(5) + (4)(-2) = -15 – 8 = -23
v · v. = (5)(5) + (-2)(-2) = 25 + 4 = 29
ac + bd
Another Dot Product example
u = <4, 6> v = <5,– 2> w = <-3, 4>
Find: u • (v + w)
<4, 6> • <5–3, -2 + 4>
<4, 6> • <2, 2>
(4)(2) + (6)(2) 8 + 12 20
v = <a, b> w = <c, d>
2121 bbaawv ac + bd
Dot Product Properties
u • v = v • u u • (v + w) = u • v + u • w
(u + v) • w = u • w + v • w (ku) • v = k(u • v) = u • kv
0 • u = 0 u • u = |u|2
Finding the angle between vectors with an alternative dot product formula
• If v and w are two nonzero vectors and is the smallest nonnegative angle between them, then
coswvwv
wv
wvand
wv
wv
1coscos
Example: Find the angle between v = <3, 4> and w = <2, 1>
Dot Product AnglePositive Acute0 RightNegative Obtuse
v • w = 6 + 4 = 10||v|| = (9 + 16) = 25 =5||w||=(4 + 1) = 5
= cos –1 10/[(5)(5) ] = cos –1 .894427191 = 26.57 degrees
You try one: Find the angle between u = <2, -6> and v = <6, 2>
Parallel and Orthogonal Vectors
• Two vectors are parallel if the angle between them is 0 or 180°
• Two vectors are orthogonal if the angle between them is 90° If the dot product of two vectors is 0 then the vectors are orthogonal.
Example: v = <3, -2> and w = <3, 2>
Are v and w orthogonal?
Check the dot product: v • w = (3)(3) + (-2)(2)
9 + -4
5
Since the dot product is NOT zero, these vectors are NOT orthogonal.
•Find the magnitude of the equilibrant of forces of 54 newtons and 42 newtons acting on a point A, if the angle between the forces is 98°. Then find the angle between the equilibrant and the 42-newton force. Equilibrant is –v. & ||-v|| = ||v||
Application: Equilibriant
Law of cosines
In parallelograms•Opposite angles : Equal•Adjacent angles : Supplementary
Angle α = 180 - <CAB.
Use Law of Sines to find < CAB.
Find the force required to keep a 2500-lb car parked on a hill that makes a 12° angle with the horizontal.BA : force of gravity.
BC: force of weight pushing against hill
BF: force that would pull car up hill
BA = BC + (–AC)
Application: Forces
BF & AC are equal, so, |AC| gives the required force.
BF and AC are parallel, BA acts as transversal; < EBD = < A.
<C and <EDB are right anglesThus, Triangles CBA & DEB have 2 corresponding angles =>are similar triangles. Therefore, the measure of angle ABC equals the measure of angle E, which is 12°.
A 520 lb force will keep
the car parked on the hill.