Thursday, November 15, 2012

TRIGONOMETRY: Sine law, Law of cosines, pythagorean identities, trigonometric identities, double angle formulas, reduction formulas, power-reducing formulas, common right triangle combination sides


Below are the most important and most useful formulas in Trigonometry which have so many applications in various fields of study, discipline and professions:

1. Sine Law (Law of Sines)
2. Cosine Law (Law of Cosines)
3. Pythagorean Theorem
4. Pythagorean Identities, Twice angle formulas
5. Double angle formulas
6. Power - reducing theorem
7. Radian measure, Reduction formulas

let:

t, theta = angle measure

a = shortest side of triangle

b = medium side of triangle

c = longest side of triangle

A = angle opposite a

B = angle opposite b

C = angle opposite c



Right Triangles


45 x 45:

a = b = 1

c = sqrt 2 = 1.4


30 x 60:

a = 1

b = sqrt 3 = 1.7

c = 2


Common Right Triangle combinations


3,4,5 ---> a=3, b=4, c=5

5,12,13 ---> a=5, b=12, c=13
 

Sine Law (Law of Sines):


MATHEMATICS: Statistics, Motion, Fencing


STATISTICS:

Mode, Median, Mean

Mode:

Rearrange the data in ascending order, the mode is occuring MOST FREQUENT

example: 5, 7, 8, 8, 8, 9  ---> mode is 8


Mean:

Mean (arithmetic mean) is the Average


Median:

odd median ---> middle

example: 5, 6, 7, 8, 9  ---> median is 7


even median ---> average of 2 medians

example: 5, 6, 8, 9

n = 4 (even number of terms)

there are two medians ---> 6 and 8

to find the median, get the average of 6 and 8


median = (6 + 8)/2 = 14/2 = 7


MOTION:

A train travels at constant speed 70 mph from Station1 to Station2. The distance between the stations is 1000 miles. Find the expression for the distance as a function of time.


solution:

Distance diminishes as it travels from station1 to station2

S = v*t

S = 70t ---> distance travelled for a certain time t


Expression:

S(t) = 1000 - 70t  ---> this is the distance as a funtion of time (the distance for a specific value of time)


where:

S = distance travelled

S(t) = distance as a funtion of time as the train travels from station1 to station2

v = constant speed or velocity

t = time of travel


FENCING:

A farmer wants to fence a length of 100 ft. Determine the number of posts required if the posts are spaced 25 ft apart.


find:

P = L/s + 1  ---> number of posts required


given:

L = 100 ft, length of fence

s = 25 ft, spacing between posts


solution:

P = L/s + 1

P = 100/25 + 1

P = 4 + 1

P = 5

GEOMETRY: Perimeter, Circumference, Largest Area, Square, Circle, Fencing, Differential Calculus, Maxima, minima


1. If 400 ft of fence is to be used, which shape generates the larger area, a circle or a square?


SQUARE:

A = s^2

using 400 ft of fence, each side of the square should be

s = 100 ft

A = s^2

A = (100)^2

A = 10,000


CIRCLE:

C = 2 * pi * r

400 = 2 * pi * r

r = 400/(2 * pi )

r = 63.7

A = pi * r^2

A = pi * (63.7)^2

A = 12,740


Comparing the areas of square and circle,

CIRCLE ---> larger area


2. What are the dimensions of a triangle of maximum area that can be inscribed in a circle such that one side of the triangle passes through the center of the circle.


solution:

let

b = base of triangle

h = height of triangle

t = angle between the hypotenuse C and the height h

C = hypotenuse = 2r



A = 1/2 b * h  ---> equation1

h = C * cos t

b = C * sin t

substituting b and h in equation1

A = (1/2) C^2 cos t sin t  ---> equation2


from double angle formulas:

sin(2t) = 2 sin t cos t

cos t sin t = (1/2)sin(2t)  ---> equation3


substituting equation3 in equation2

A = (1/4) C^2 sin (2t)


using the double angle form,

A = (1/4) C^2 (2 cos t sin t)


derivative of A with respect to t ---> derivative of a product

dA/dt = (1/4) C^2 (2 cos t sin t)

dA/dt = (1/4) C^2 [ 2 (cos t cos - sin t sint) ]

dA/dt = (1/4) C^2 [ 2 (cos^2 t - sin^2 t) ]


from double angle formulas:

cos^2 t - sin^2 t = cos 2t


substituting

dA/dt = (1/4) C^2 [ 2 (cos 2t) ]

dA/dt = (1/4) C^2 * 2 cos 2t


to find the maximum area, equate dA/dt = 0

dA/dt = (1/4) C^2 * 2 cos (2t) = 0


as the angle between the hypotenuse and the adjacent side(h) approaches 90, cos t approaches zero because the adjacent side (h) gets shorter and shorter while the opposite side (b = base of triangle) gets longer and longer

sin 90 = 1

cos 90 = 0


for dA/dt = 0

2 * t = 90

t = 45 degrees


substituting the values for C = 2r and t = 45

h = C * cos t

h = 2r * cos 45

h = 2r * sqrt(2)


b = C * sin t

b = 2r * sin 45

b = 2r sqrt(2)


The area is maximum when t = 45 degrees

and

b = h = 2r sqrt(2)

which is an  ISOSCELES right triangle.

Wednesday, November 14, 2012

GEOMETRY: Interior, Exterior angles of any polygon, ratio of the angles


Interior angles of any polygon

Ia = 180(s - 2)


where:

Ia = sum of all interior angles of any polygon

s = number of sides of the polygon


for triangles: Ia = 180

for rectangles, trapezoids, quadrilaterals, parallelograms and any four-sided polygons: Ia = 360

for a pentagon: Ia = 540

for a hexagon (hex-bolt): Ia = 720


Exterior angles of any polygon

Ea = 360


where:

Ea = sum of all exterior angles of any polygon


1. Determine the largest angle of a Triangle if the ratio of the angles is 1:3:5


find:

x = shortest angle

5x = largest angle of triangle

solution:

1x + 3x + 5x = 180

9x = 180

x = 180/9

x = 20

3x = 3 * 20 = 60

5x = 5 * 20 = 100

IQ TEST: Math, Percentages, Markdowns, Discounts, Original, discounted Price, Ratio, Proportion


1. The ratio of two numbers is 1:4, if the smaller number is increased by 7, the ratio becomes 1:2. Find the two numbers.

find:

x = smaller number (based on 1:4 ratio)

4x = bigger number (based on 1:4 ratio)

solution:

(x + 7)/4x = 1/2

4x = 2x + 14

2x = 14

x = 7

4x = 4 * 7 = 28



2. A certain product has been marked down twice. The first was 20% and the second 25%. If the original price was $100, what is the final price after the two markdowns?

find:

x = final price


solution:

after first mark down of 20%

Price = 100 * 0.80 = 80


after second mark down of 25%

x = 80 * 0.75 = 60


3. The original price was $100. A discount of 10% has been offered, then the price went back to the original price. What is the percentage of increase with respect to the discounted price?

find:

x = percent of increase with respect to the discounted price

solution:

after 10% discount

discounted price = 100 * 0.90 = 90


to go back to original price

increase = 100 - 90 = 10


x = (increase/discounted price) * 100%

x = 10/90 * 100%

x = 1/9 * 100%

x = 11.11% ---> [ 1/9 is repeating 0.1111111111111111111111111111 ]


IQ TEST: Math, Pie, Cake, Divisions, Total, Sum, Parts, Age Problems


1. A birthday cake is to be divided and distributed such that the birthday celebrant will have a quarter share and the rest will have half as big as the celebrant's share. How many pieces was the cake divided?

find:

x = number of equal shares

solution:

TOTAL = sum of parts

1 = 1/4 + x(1/2 * 1/4)

x(1/8) = 3/4

x = 6

Cake divisions = 6 + 1 = 7


2. A school has 55 students for each teacher. If the school has a total of 2240 students and teachers, how many teachers are there?

find:

x = number of teachers

solution:

TOTAL = sum of students and teachers

2240 = 55x + x

56x = 2240

x = 40


3. Roland is twice as old as Adrienne. Twenty years ago, Roland was four times as old as Adrienne. How old is Adrienne now?

find:

A = Adrienne's age now

given:

now

A = Adrienne's age

R = Roland's age

R = 2A ---> equation 1


twenty years ago

A - 20 = Adrienne's age

R - 20 = Roland's age

R - 20 = 4(A - 20) ---> equation 2


substituting equation 1 in 2

2A - 20 = 4(A - 20)

2A - 20 = 4A - 80

2A = 60

A = 30

Saturday, November 3, 2012

IQ TEST: Math, Working together, Job, Work problems at the same rate, inverse proportions


1. John can finish the job in 3 hours while Amy can finish the same job in 4 hours. If both work on the same job, how long will they finish?


find:

T = total time if both work together

t1 = time for John to finish the job alone

t2 = time for Amy to finish the job alone

solution:

1/T = 1/t1 + 1/t2


1/T = 1/3 + 1/4

1/T = 7/12

T = 12/7 or 1.7 hours


2. Two carpenters can finish 4 cabinets in 4 hours. If the working rate is the same, how many cabinets can 7 carpenters finish in 24 hours?


find:

J2 = number of cabinets finished by 7 carpenters in 24 hours

given:

m1 = 2 carpenters

t1 = 4 hours

J1 = 4 cabinets

m2 = 7 carpenters

t2 = 24 hours

solution:

m1*t1/J1 = m2*t2/J2


2(4)/4 = 7(24)/J2

2 = 168/J2

J2 = 84 cabinets


3. A factory has an output of 100 units. An additional 20% of output has required an increase of 4 units per worker. If they are working at the same rate, determine the number of workers.



find:

x = number of workers

solution:

OUTPUT = rate * workers

additional output = additional rate * workers


100(0.2) = 4x

20 = 4x

x = 20/4

x = 5


check:

normal capacity:

output = 100

rate = 100/5 = 20 units per worker


at 20% over normal capacity (120 units):

rate = 20 + 4 = 24

output = 24 * 5 = 120 units