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7.1 & 7.2 Law of Sines ue triangle – A triangle that does not contain a right angle. C B A 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 = =

# 7.1 & 7.2 Law of Sines

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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

### Text of 7.1 & 7.2 Law of Sines

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

• Find the area of triangle DEF in the figure.

• 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.

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