## Tuesday, February 6, 2007

### 6.2 Law of Cosines

6.2 Law of Cosines
To solve an oblique triangle where you are given the measurement of the three sides (SSS) or two sides and the included angle (SAS), you would use the Law of Cosines:

Law of Cosines:

a2 = b2 + c2 - 2bc cos A

b2 = a2 + c2 - 2ac cos B

c2 = a2 + b2 - 2ab cos C

Example 1: given SSS - find the three angles of the triangle given a = 9, b = 3 and c = 11

a2 = b2 + c2 - 2bc cos A
92 = 32 + 112 - 2(3)(11) cos A
81 = 9 + 121 - (66) cos A
81 - 9 - 121 = - 66 cos A
-49 = -66 cos A
49/66 = Cos A
A = 42.06166467°

b2 = a2 + c2 - 2ac cos B
32 = 92 + 112 - 2(9)(11) cos B
9 = 81 + 121 - 198 cos B
-193 = -198 cos B
193/198 = cos B
B = 12.90352054°

c2 = a2 + b2 - 2ab cos C
112 = 92 + 32 - 2(9)(3) cos C
121 = 81 + 9 - 54 cos C
31 = -54 cos C
-31/54 = cos C
C = 125.0348148°

Example 2: Two Sides and the Included Angle - SAS
C = 108° , a = 10, b = 6.5
c2 = a2 + b2 - 2ab cos C
c2 = 102 + 6.52 - 2(10)(6.5) cos 108°
c2 = 182.4222093
c = 13.50637662

Now find the other 2 angles:
a2 = b2 + c2 - 2bc cos A
102 = 6.52 + 13.506376622 - 2(6.5)(13.50637662) cos A
100 = 224.64722093 - 175.582896 cos A
-124.6722093 = - 175.582896 cos A
.7100475736 = cos A
A = 44.76121382°

B = 180° - 108° - 44.76121382° = 27.23878618°

Example 3: If a street light is 2 feet in length on the post, the top is 3 feet and the distance from the bulb to the post is 4.5 feet. What is the angle measurement from the post to the top of the sign?

4.52 = 22 + 32 - 2 (2)(3) Cos A
20.25 = 4 + 9 - 12 Cos A
7.25 = -12 Cos A
-7.25/12 = Cos A
A = 127.1688997°
so therefore, the angle from the post to the top of the sign would be
180° - 127.1688997° = 52.83110034°

Heron’s Area Formula - given any triangle with sides of lengths a, b, and c, the area of the triangle is:

Given the lengths of the sides a, b, and c and the semiperimeter s,

Example 4: The Landau Building in Cambridge, Massachusetts, is a building with a triangular-shaped base. The lengths of the sides of the triangular base are 145 feet, 257 feet, and 290 feet. Find the area of the base of the building.

s = (145 + 257 + 290)/2 = 692/2 = 346 feet

Area = √(346(346 - 145)(346 - 257)(346 - 290))

= 18617.66 square feet
OR
Using Area = ½ bc Sin A, first find the measure of angle A

a2 = b2 + c2 - 2bc sin A
1452 = 257 2 + 290 2 - 2(257)(290) cos A
21025 = 66049 + 84100 - 149060 Cos A
-129124 = - 149060 cos A
129124/149060 = cos A
A = 29.97365689°

Area = ½ (257)(290) sin 29.97365689°
Area = 18617.66 square feet

## Monday, February 5, 2007

### 6.1 Law of Sines

6.1 Law of Sines

A. Solving oblique triangles - triangles that have no right angles.

To solve an oblique triangle, you need to know the measure of at least one side and any two other parts of the triangle - that is, two sides, two angles, or one angle and one side.
1. Two angles and any side (AAS or ASA)
2. Two sides and an angle opposite one of them (SSA)
3. Three sides (SSS)
4. Two sides and their included angle (SAS)

The first two cases can be solved using the LAW OF SINES, whereas the last two cases require the Law of Cosines.

Here is a great web site to check out: http://hyperphysics.phy-astr.gsu.edu/hbase/lsin.html
and http://www.algebralab.org/studyaids/studyaid.aspx?file=Trigonometry_LawSines.xml
and http://www.ies.co.jp/math/java/trig/seigen/seigen.htmltp://

Check these out!

Law of Sines
If ABC is a triangle with sides a, b, and c, then

a/ (sin A ) = b/ (sin B) = c/ (sin C )

or

(sin A)/ a = (sin B)/ b = (sin C)/ c

This triangle now looks like the picture below.

Remember that the side and angle of a triangle that share the same name
are always across from each other.
In order to set up an equation using the sine function, we have to create a right angle.
Construct a height segment in the triangle by dropping a perpendicular segment
from angle C to side c. This triangle now looks like the picture below.
 Using the smaller triangle on the left that includes angle A and sides b and h,we can set up an equation involving sine. Using the triangle on the right half that includes angle B and sides a and h,we can set up and equation involving sine. Both of these equations involve “h”.Solve both equations for “h”. Set the two expressions for “h” equal to each other. Divide both sides by ab. Reduce each fraction. Final equation that uses the sine function for oblique triangles.

you can do this process again to get (sin A)/a = (sin C)/c

Example 1: Use the given information to solve the triangle:

B = 45°, c = 15, C = 120°

A = 180° - 45° - 120° = 15°

(Sin B)/ b = (Sin C)/c

Sin 45° / b = Sin 120° / 15

b = 12.24744871

(Sin A)/a = (Sin C)/c
sin 15° / a = sin 120° / 15

a = 4.482877361

Recall from Geometry that the largest angle is opposite the longest side.
A = 15° B = 45° and C = 120°
a = 4.48 b = 12.25 and c = 15
so yes it makes sense.

Check out this web site:
http://www.sparknotes.com/math/trigonometry/solvingobliquetriangles/section3.rhtml

The Ambiguous Case (SSA)
Consider a triangle in which you are given a, b, and A (h = b sin A).

I. Angle A is acute

A. If angle A is acute and a is less than h, then it is impossible to make a triangle.
B. If angle A is acute and a = h, then there is one right triangle.
C. If angle A is acute and a is greater than b, then the possible number of triangles is one.
D. If angle A is acute and h is less than a which is less than b, then there is a possibility of two triangles.

II. Angle A is obtuse

A. If angle A is obtuse and a is less than or equal to b, then it is impossible to make a triangle.
B. If angle B is obtuse and a is greater than b, then there is only one possible triangle.

Because the sine function is positive in both the I and II quadrant, we have to check to see if there is one or two possible solutions using the law of sines.

Example 2:
Given A = 58°, a = 4.5 and b = 5, find all possible solutions.

Sin 58°/ 4.5 = sin B/ 5

Sin B = .9422756624

B = 70.43730473°

C = 180° - (58° + 70.43730473° )= 180°-128.43730473° = 51.56269527°

then Sin 58°/ 4.5 = sin C/ c

c = 4.156367925

OR if 70.43730473° is the reference angle then

180° - 70.43730473° = 109.5626953° = B

C = 180° - (58° + 109.5626953° )= 180°-167.562693° = 12.43730473°

Sin 58°/ 4.5 = sin C/ c
c = 1.142824717

A = 58°, B = 70.43730473°, C = 51.56269527°
a = 4.5, b = 5, and c = 4.156367925

Or

A = 58°, B = 109.5626953° , C = 12.43730473°
a = 4.5, b = 5, and c = 1.142824717

Since both of these make sense, there are 2 solutions.

Example 3:
Given A = 110°, a = 125, and b = 100, solve the triangle.

(Sin 110°)/ 125 = (Sin B) / 100

B = 48.74255246°
C = 180° - (110° + 48.74255246°) = 21.25744754°

(Sin 110°)/ 125 = (Sin 21.25744754°) / c
c = 48.22842776

OR if B = 48.74255246° was the reference angle then
B = 180° - 48.74255246° = 131.2574475°

so C = 180 - (110° + 131.2574475°) = -61.25744754 which is impossible
so there is only one solution.

Area of an Oblique Triangle:

The area of any triangle is one-half the product of the lengths of two sides times the sine of their included angle. That is:
Area = ½ bc sin A = ½ ab sin C = ½ ac sin B

Example 4:
Find the area given: B = 74°30' , a = 103 m, c = 58 m
Area = ½ (103)(58)(sin 74°30' ) = 2878.364164 square meters

6.1 homework #49; page 434; #1 - 33 odd, 34, 38, 41

## Saturday, February 3, 2007

### Precalculus 5.5 Multiple-Angle and Product-Sum Formulas

5.5 Multiple-Angle and Product-Sum Formulas

Double Angle

A. Evaluating Functions Involving Double Angles:
Example 1:
Given: tan θ = 3/4, find cos 2θ

tan θ = opp/adj so therefore using Pythagorean theorem,
we know the hypotenuse is 5, so cos θ = 4/5

cos 2θ = 2 cos2θ - 1 = 2 (4/5)2 - 1 = 2 (16/25) - 1 = .28 = 7/25

B. Solving a Multiple-Angle Equation
Example 2:
sin2x + cos x = 0

2 sin x cos x + cos x = 0

cos x (2 sin x + 1) = 0

Solving for x:

cos x = 0

x = π/2 , 3π/2

2 sin x + 1 = 0

sin x = - ½

x = 7π/6 and 11π/6

Therefore the general solution is

x = π/2 + π n and x = 7π/6 + π n

C. Power-Reducing Formulas

Power-Reducing/Half Angle Formulas

Example 3:

cos4 x = (cos2x)2

Product-to-Sum Formulas
sin u sin v = ½ (cos (u - v) - cos (u + v))
cos u cos v = ½ (cos (u - v) + cos (u + v))
sin u cos v = ½ (sin (u + v) + sin (u - v))
cos u sin v = ½ (sin (u + v) - sin (u - v))

Sum-to-Product Formulas

check out this web-site for additional help:
http://wps.prenhall.com/esm_blitzer_algtrig_2/0,7303,911519-,00.html

### Precalculus 5.4 Sum and Difference Formulas

5.4 Sum and Difference Formulas

Sum and Difference Formulas

sin (u + v) = sin u cos v + cos u sin v
sin (u - v) = sin u cos v - cos u sin v

cos (u + v) = cos u cos v - sin u sin v
cos (u - v) = cos u cos v + sin u sin v

A) Evaluating a Trigonometric Function:
1. Example 1

sin 75° = sin (30° + 45°)
sin 75° = sin 30° cos 45° + cos 30° sin 45°

Example 2: sin 90° = 1

sin 90° = sin (30° + 60°)
sin 90° = sin 30° cos 60° + cos 30° sin 60°

B) Proving a Cofunction Identity

Example 3:

cos (π - θ ) + sin (π/2 + θ ) = 0

cos π cos θ + sin π sin θ + sin (π/2) cos θ + cos (π/2) sin θ = 0

(-1) cos θ + (0)(sin θ ) + (1) cos θ + (0) sin θ = 0

0 = 0

Example 4:
(Cos ( x + h) - cos x)/h = (cos x (cos (h) - 1))/h - (sin x sin h)/h

(Cos x cos h - sin x sin h - cos x)/ h = (cos x (cos (h) - 1))/h - (sin x sin h)/h

(Cos x ( cos (h) - 1) - sin x sin h)/h = (cos x (cos (h) - 1))/h - (sin x sin h)/h

(cos x (cos (h) - 1))/h - (sin x sin h)/h = (cos x (cos (h) - 1))/h - (sin x sin h)/h

C) Solving a Trigonometric Equation:

Example 5:

cos (x + π/6) - cos (x - π/6) = 1

cos x cos π/6 - sin x sin π/6 -(cos x cos π/6 + sin x sin π/6) = 1

cos x cos π/6 - sin x sin π/6 - cos x cos π/6 - sin x sin π/6 = 1

-2 sin x (½) = 1
-1 sin x = 1
sin x = -1

x = 3π/2

5.4 Homework #46: pg 408; # 3, 5, 11, 15, 19 - 27 odd, 35 - 57 odd