Study Online Factors

In online study on factors, it is essential to know the definition of factors.The set of numbers which produce the remainder zero by dividing a particular number. These sets of numbers are called as factors. We can express a number by the multiples of another two numbers called as factors.

For Example,

                   10= 2 x 5

                   32=4 x 8

     Here 2, 5 are the factors of 10 and 4, 8 are the factors of 32.

     Types of Factors and Sample problems:


For the online study on factors, knowing the types of factors is an essential one. The types of factors for online study are as follows:

There are two types of factors,

Prime Factors
Composite factors

Online study on Prime Factors:
     The numbers that can be expressed as the multiple of one and that number itself are called as prime numbers. The factors of prime numbers are called as prime factors.

Consider the following Examples,

3= 1 x 3 (1, 3 are the prime factors of 3)

5= 1 x 5 (1, 5 are the prime factors of 5)

7= 1 x 7 (1, 7 are the prime factors of 7)

Online study on Composite Factors:

     The factors that are not a prime numbers are called as composite factors.

Consider the following example,

24=4 x 6

Here the numbers 4, 6 are not a prime number. Therefore, 4, 6 are called as composite factors of 24.

Example:

Find the factors of 130.
Solution:

130= 1 x  130

130= 2 x  65

130= 5 x  26

130= 10 x 13

So the factors of 130 =1, 2,5,10,13,26,65,130.

Important note: We can express all the numbers except “1” as a product of prime numbers are called as prime factorization.

Consider the following Example,

24 = 4 x 6

     = (2 x 2) x (2 x 3)

24 = 23 x 3

Here 2, 3 are the prime numbers. Hence, we conclude that all the numbers can be expressed as a product of prime numbers.

Prime factorization tree:

     By using the prime factorization tree, we can get all the prime factors of the given number.

Example:

Find the prime Factors of 124
Solution:

                                       124

                                         /  \

                                     2     62

                                             /  \ 

                                          2   31

                            124 = 2 x 2 x 31


Practice Problems on factors:


Following are the practice problems given for online study on factors.

1. Find the factors of 524.

2. Find the Prime factors of 100

3. Find the composite factors of 1056.

Histogram Calculation

A histogram is a graphical representation of frequencies of a variable. The variable range is equally divided and is scaled on the x axis and is continuous. The frequencies of the variable in each range is plotted on the y-axis and a rectangle is drawn  for each interval with the height of the rectangle equal to the frequency of the range and the width representing the range. The width of the rectangles in each range will be equal since total range is equally divided into intervals. The walls of adjacent intervals share the same line.The height of the histogram make the interpretation of the frequencies of each interval easier compared to a tabular representation. The more the height of the rectangle in an interval the more is the frequency within the interval.

Parts of Histogram:

Title:

The title section in the histogram briefly explains the information about the variable that is used in the graph.

Bars:

The bars are explained by thier height and width. The height represents the frequency of observations falling into the interval. The width represents the interval.

Legend:

The legend provides extra information about the relation to the documents where the data came from and how the dimensions were gathered.

Horizontal X-Axis:

The horizontal X-axis provide the scale value, which represent the dimension that fit into the data. These dimensions were usually recognized to the periods. Plot the horizontal X-axis points with in the bar chart which respect the values of Vertical Y-axis.The dimension determine the interval to be fixed.

Vertical Y-Axis:

The vertical Y axis represent the scale of frequencies and is optimized to represent the data on the graph. The units are acquiesced and is linear starting from zero.

Histogram calculation - Conditions in Histograms:

The scale should be supposed to have all the data values. The scale divides the range into equal parts.


Steps to construct histograms:

• Classify the data of the given histogram
• Identify the time period for the data
• Tabulate the data for the given histogram.
• Establish the range of the given data
• come to a decision for the number of height and width of each bar
• Count the number of items in each bar.
• Make a bar chart using the data given.

Histogram calculation - Example Problems:

Histogram calculation - Problem 1:

Calculate the histogram for the given data.

5    10    15    20    25    30    35    40    45    50    55    60
2      4      7    10      5      7      7      4      4    10      7      9



Histogram calculation - Problem 2:

calculate the histogram for the following given data.

100-200    200-300    300-400    400-500    500-600    600-700
45             34             89              23            67               98



Histogram calculation - Problem 3:

calculate the histogram for the given data set.

1   2      3      4      5    6      7      8      9    10
7    3    32    56    15    4    78    34    45    56

Frequency Distribution Tables

Frequency distribution:

The frequency distribution represents the number of observations with in the interval These are  graphically or tabular format.But the intervals must be mutually exclusive and exhaustive,and these are used with in a statistical context.The frequency distribution is a table it contains classes and its representing frequency.The class means quantitative or qualitative type in which the placed of raw data.The data is two types .1)Primary data 2)Secondary data.Primary data is collected by who are the investigator,some times an investigator uses another investigator of primary data.This type of data is called secondary data.The secondary data grouping and presenting in form of table.These type of tables are called Frequency  distribution tables.The data form is in the table shows below.

Data Range:

It is Highest value-lowest value

Class width

Range/ desired number of classes

Upper/lower class limit – upper/lower class Limit of next class

Upper class boundary – lower class Boundary

Class midpoint

Xm   = ( Lower bound + Upper bound)/2

Or

Xm = (Lower Limit + Upper Limit) /2


Frequency Distribution table:


.We can represents The frequency distribution table in number of ways.

1)Group frequency table or graph(polygon,chart)

2)Regular frequency table or Bar graph(Histogram)

When the data is from nominal or ordinary scale then we use Bar graph.

When the data measurements consists more categories than the listed in regular table then we use Group frequency distribution.


Important Notes:


1)The grouped frequency distribution table must use when the range of scores is large, causing a regular frequency table to have too many entries in the score categories (X column). The guidelines for a group frequency table include   approximately 10 rows in the table

2)  The Interval width of 2,5,10,20,50,100 should be used  it depends on the number of rows chosen

3)  Each interval of the  first (lowest) value of should be a multiple of the interval width

4) Note all intervals without missing any, The top interval must contain the highest observed X value and the bottom interval should contain the lowest observed X value.

Example:

For  the given data below into a frequency distribution table and show percentage of each category.
8, 9, 8, 7, 10, 9, 6, 4, 9, 8,
7, 8, 10, 9, 8, 6, 9, 7, 8, 8



x    f    ρ
10    2    0.1
9    5    0.25
8    7    0.35
7    3    0.15
6    2    0.10
5    0    0
4    1    0.05

Solving Irrational

solving irrational numbers:

Irrational numbers in difference to rational numbers aren’t presented a fraction of the shape: m / n,  where m and n are integers. There are numbers of a new type, which are calculated with any accuracy, but can’t be changed by a rational number. They can appear as outcome of geometrical measurements,

for example:

- a ratio of a square diagonal duration to its side length is equal to √2.

- a ratio of a circumference duration to its diameter length is an irrational number  Pi



Where can you find special irrational numbers?(solving irrational)


The answer to this depends on what you consider 'special'.  Mathematicians have proved that definite special numbers are irrational, for example Pi and e.  The number e is the base of natural logarithms.   It is irrational, just like Pi, and has the value 2.718281828459045235306....

It's not easy to just "come up" with such special numbers.  But you can easily discover more irrational numbers after you've found that most square roots are irrational.  For example, what do you think of √2 + 1?  Is the result of that addition a rational or an irrational number?  How can you know?  What about other sums where you add one irrational number and one rational number, for example √5 + 1/4?

You can also add two irrational numbers, and the amount will be many times irrational.  Not always though; for example, e + (-e) = 0, and 0 is rational even though both e and -e are irrational.  Or, take 1 + √3 and 1 - √3 and add these two irrational numbers - what do you obtain?

Or, increase/separate an irrational number by a rational number, and you get an irrational number.  For example, √7/10000 is an irrational number that is moderately close to zero.  Yet another possibility to discover irrational numbers is to multiply square roots or other irrational numbers.  Sometimes that outcome in a rational number though (when?).  Mathematicians have also calculated what happens if you raise an irrational number to a rational or irrational power.

Yet more irrational numbers begin when you take logarithms, or calculate sines, cosines, and tangents.  They don't have any particular names, but are just called "sine of 70 degrees" or "base 10 logarithm of 5" etc.  Your calculator will give you decimal approximation to these.

Examples of Irrational Number


`Pi`,  √10 Are few examples of irrational numbers.

Solved Example on Irrational Number

Identify the irrational number.

Choices:
A. 0.3
B. 3 / 5
C. √6
D. √25
Correct Answer: C

Solution:
Step 1: The value of  is 2.4494897427831780981972840747059...
Step 2: The value of  is 5.
Step 3: The value of  is 0.6.
Step 4: Among the choices given, only  is an irrational number

FRACTIONS UNLIKE DENOMINATORS

A fraction  is a number that can represent part of a whole. An example is 3/4, in which the numerator, 3, tells us that the fraction represents 3 equal parts, and the denominator, 4, tells us that 4 parts make up a whole.

How to Add Fractions with Unlike Denominators:

To add fractions containing unlike quantities (e.g. quarters and thirds), it is necessary to convert all amounts to like quantities. It is easy to work out the chosen type of fraction to convert to; simply multiply together the two denominators (bottom number) of each fraction.

Consider adding the following two quantities:

3/4 + 2/3

First, convert  3/4  into twelfths by multiplying both the numerator and denominator by three3/4 x 3/3 =9/12

Note that  3/3 is equivalent to 1. which shows that  ¾ is equivalent to the resulting  9/12.

 Secondly, 2/3convert  into twelfths by multiplying both the numerator and denominator by four:2/3 x 4/4 = 8/12. Note that 4/4  is equivalent to 1, which shows that  2/3  is equivalent to the resulting 8/12 .

9/12 + 8/12 = 17/12.


Simplify the Fraction

Example: Find the Sum of 2/9 and 3/12

Determine the Greatest Common Factor of 9 and 12 which is 3

Either multiply the denominators and divide by the GCF (9x12=108, 108/3=36)

OR - Divide one of the denominators by the GCF and multiply the answer by the other denominator (9/3=3, 3x12=36)

Rename the fractions to use the Least Common  Denominator(2/9=8/36, 3/12=9/36)

The result is 8/36 + 9/36

Add the numerators and put the sum over the LCD = 17/36

Simplify the fraction if possible. In this case it is not possible.

Example:

Find. 3/4  - 2/3

Solution:

Since the fractions have different denominators, they cannot be subtracted until they have the same denominators.

We can express both fractions as twelfths.

3/4 = 4/12 and 2/3 = 8/12

9/12 - 8/12 = 1/12.

Example:

Find

 5/6 + 2/3

Solution:

Since the fractions have different denominators, they cannot be added until they have the same denominators.

We can express both fractions as sixths.

2/3 = 4/6

5/6 + 3/6 = 8/6

Graphing Inequalities

In this article we are going to discuss about graphing inequalities concept.A coordinate graph is called as the Cartesian coordinate plane. The graph contains a couple of the vertical lines are called coordinate axes. The vertical axis of the y axis value and the horizontal axis value is the x axis value. The points of the intersection of those two axes values are called the origin of coordinate graphing pictures. In this graph linear equation is in the form of y = mx + c. I

Graph of inequalities

Below you can see some examples on graphing inequalities -

Problem 1 :

Solving the system of inequalities for two variables equation use graphing method 7x – 6y < 5 and draw the graph for a given inequalities.

Solution:

We find the plotting points of the given inequality equation. The given equation is

            7x - 6y < 5

We are going to find out the plotting points for a given equation. In the first step we are going to change equation in the form of y = mx + c, we get the following term

                7x - 6y < 5                  

                  7x < 6y + 5

In the next step we are multiply the value -1 to equation (1), we get

            -7x > - 6y - 5

               6y > 7x – 5

In the above equation we put x = -1 we get

            6y > -7 - 5

            y > -2

In the above equation we put x = 2 we get

            6y > 14 – 5

             y > 1.5

We draw a graph for given equation. We get





Solving the system of inequalities for two variables equation use graphing method 4x – 3y > 2 and draw the graph for a given inequalities.

Solution:

We find the plotting points of the given inequality equation. The given equation is

            4x - 3y > 2

We are going to find out the plotting points for a given equation. In the first step we are going to change the above equation in the form of y = mx + c, we get

      4x - 3y > 2

            4x > 3y + 2                 

In the next step we are multiply the value -1 to a above equation, we get

               -4x < -3y - 2

                   3y < 4x - 2

In the above equation we put x = -1 we get

            3y < -4 - 2

             y = -2

In the above equation we put x = 2 we get

            3y < 8 - 2

             y < 2

We draw a graph for given equation. We get

The graphical concept of x intercept and y intercepts is Very simple. The x intercepts are the points where the graph crosses the x-axis, and the y-intercepts are the points where the graph crosses the y-axis. The problems start when we deal with intercepts algebraically.

The X and Y intercept are drawn in the graph to find the slope of the graph from the given equation.

Then, algebraically,

an x intercept is a point on the graph where the point y =  zero, and
A y intercept is a point on the graph where the point x = zero.

Procedure for finding X and Y intercept

x Intercept

Step1: For finding the X intercept we need to arrange the given equation in the slope intercept form.

Step2: The slope intercept form is    y = mx + c

Step3: After this we substitute y=0.

Step4: The equation becomes mx + c = 0, now we need to add/subtract constant C on both sides.

Step5:  Then divide the constant by the slope.  x = c/m.

y Intercept

Step1: For finding the y intercept we need to arrange the equation in slope intercept form.

Step2: After this, substitute x = 0

The equation becomes y= c, this is the Y intercept.


y and x intercept for 2x+y=8.


2x+y=8

The given equation is in the form of slope intercept form i.e. Y = mX + C

To find X Intercept, put y = 0

0 = -2x + 8

Subtract 8 on both sides,

0 – 8 = -2x + 8 – 8

-2x = -8

Divided 2 on both sides

-2x/2 = `-8/2`

-x = -4

Here the x intercept is 4.

To find the Y Intercept, put x = 0

y = 0 + 8

y = 8

Here the y Intercept is 8.

Using this we can plot the graph.

Vector Components

Here, we are going to discuss about the components of vector, generally vector has both the direction and the magnitude. Components, we mean as the parts of vectors, otherwise we can tell the two perpendiculars that added together means the components of the vector will give the original vector .The concept of the component vector is also the same as the rule of the vector addition .Vectors are denoted by the arrow look like as the below`->`

Rules of Vector Components

Here, we are following some rules in the components of the vectors those are given below:

• Components should always be perpendicular sometimes we called, the orthogonal components.
• The component s of the vector may be in any axis (x and the y axis) we called the horizontal or the vertical dimension.
• The direction of the components is look like the head to tail, so that we can add that vector. (because it is the components)
• So the components of the vectors are mutually independent .It will be shown through the following figure.
• If we are adding those x and y vectors we can get the resultant vector.

Examples on components of vectors

Example 1: Calculate the components of vectors along with the coordination of the position vector of P(− 4, 3)

Solution: Here, the position vector is given below `vec(OP)` = −7`veci` + 13`vecj` ,So the Component of the position vector `vec(OP)`  along x-axis is given as  − 7`veci ` . the component of  the position vector `vec(OP)`along x-axis is the  vector with the  magnitude of the 7 and here, we have to mention the direction , its direction is along the negative direction of x-axis ,because -7 is given.

So, the Component of the position vector  `vecOP` along  the y-axis is 13. The component of the position vector which as `vec(OP)` along y –axis is a vector with the  magnitude  of 13,here the direction of the vector have the positive. So,it will be the positive direction along the y-axis.

Example 2: Calculate the x-component of  d2  using the sin function the opposite value is given as 35 degree and the hypotenuse value is 33m?

Solution : Sin `theta` = opposite/hypotenuse

Sin 35 = d2x/23m

d2x = (33m)(sin 35)

d2x = (33m)(0.5736)

d2x = 18.9288m

We can check that the value, whether it is positive or negative directions. In this problem This x-component points to the right so, it is positive.

d2x = 18.928756m is the answer.

Bar Graph Examples

Students can learn about Bar Graphs from the Bar Graph Examples. They can get help with plotting bar graphs from the online Statistics tutors.

A bar chart or bar graph is a chart with rectangular bars with lengths proportional to the values that they represent. The bars can also be plotted horizontally.

Bar charts are used for plotting discrete (or 'discontinuous') data i.e. data which has discrete values and is not continuous. Some examples of discontinuous data include 'shoe size' or 'eye color', for which you would use a bar chart. In contrast, some examples of continuous data would be 'height' or 'weight'. A bar chart is very useful if you are trying to record certain information whether it is continuous or not continuous data.


Examples of bar graphs


Definition for bar graph:

Bar graph can be either vertical or horizontal. There may be just one bar or more than one bar for each interval. Sometimes each bar is divided into two or more parts. The data is made up of more than one category. In this section, you will work with a variety of bar graphs. Be sure to read all titles, keys, and labels to completely understand all the data that is presented. The Bar graph examples cover all the types of bar graphs.

Examples of bar graphs:

Bar graph Examples 1:

The results for last year high school math grades have been calculated. How many more 9th graders made A’s in math than 11th graders?

Solution:



Step1: Looking under the category of A’s, locate the graph for 8th graders. The interval on the left side of the bar graph reads 75 at the highest points on the graph for 8th graders. That means 75 8th graders made an A in math class.

Step 2:  Now do the same thing in step 1 again except the time instead of looking at the graph for 8th graders. We look at the graph for 10th graders. The interval on the left side of the bar graph reads 50 at the highest point on the graph for 10th graders. That means 50 10th graders made an A in math class.

Step 3:  Subtract the 50 10th graders from the 75 8th graders. There is an excess of 25 students. This means that 25 more 8th graders made on A in math than 10th graders.


Example for bar graphs 2:


Bar graph Examples 2 :

Graph show the number of rainy days in the months since May to September. How various rainy days are here in July, August, and September altogether?

Solution:



Find the number of rainy days in every of the three months and add them alone

Step 1:            Find the months scheduled along the horizontal axis.

The months are scheduled in a row from May to September.

Step 2:           Locate the bars for July, August and September.

Step 3:           Find the number of rainy days for July, August and September. Move up the bar from the label “July” until you approach to the top of the bar.  After that move to the left to discover the number of rainy days. The answer is 7, since the top of the bar is halfway among the 6 and 8. Do the same for August and September.

July – 7 rainy days

August – 3 rainy days

September – 6 rainy days

Step 4:           Add the three months:

7 + 3 + 6 = 16 rainy days

There are 16 rainy days in July, August and September.

Students can also avail help with Statistics homework problems involving Bar graph examples from the online tutors.

Formula for solving quadratic equations

A quadratic functions in the variable of x is an equation of the general form ax^2 + b x + c = 0, Where a, b, c are real numbers, a not equal to Zero. 2x^2 + x – 300 = 0 is a quadratic equation.

The simplest way to solve ax2 + bx + c = 0 for the value of x is to factor the quadratic, set each factor equal to zero, and then solve each factor. But sometimes the given quadratic equation is too messy, or it doesn't factor at all. Factoring may not always be successful; the Quadratic Formula always helps to find the solution.

The Quadratic equation form is ax2 + bx + c, where a, b, and c are just numbers; they are known as the numerical coefficients. The Formula is derived by the process of completing the square.

That is, ax2+ bx + c = 0, a not equal to Zero, is called the standard form of a quadratic function


Formula for Solving Quadratic Equation:


The solutions of any quadratic equation, ax2 + bx + c = 0 is given by the following formula, called the formula of quadratic equation:

X = ` ( -b +- sqrt(b^2 - 4ac))/(2a) `

For using the Quadratic Formula to work, the equation must be re arranged in the form (quadratic) = 0. Also, the 2a in the denominator of the Formula is underneath everything above in the numerator, not just the square root. And there 2a  is under there, not just a plain 2.


solving quadratic equation - Example problems:


solving quadratic equation -  problem 1:

x2  + 9x + 8  = 0

Solution:

Factoring method is used to solve for x,

x2 + 1x + 8x + 8 = 0

x(x + 1) + 8(x + 1) = 0

(x + 1) ( x + 8 )=0

X = -1, -8

The answers are -1,-8.

solving quatratic equation -  Problem 2:

Solve: x2 + 8x + 12 = 0.

Solution:

Factoring method is used to solve for x,

x2 + 5x + 3x + 12=0

x(x+5) + 3(x + 4) =0

(x+5)(x+3)=0

x = - 5 , - 3

The answers are  -5, -3.

solving quatratic equation -  Problem 4:

Solve 3x2 + 5x = -1 for x.

Solution: First find the standard form of

The equations and determine a, b,and c.

2x2 + 6x + 1 = 0

a = 2

b = 6

c = 1

Plug the values you found for

a, b, and c into the

Quadratic formula.

X = ` ( -b +- sqrt(b^2 - 4ac))/(2a) `

Perform any indicated operations.

X = ` ( -6 +- sqrt(6^2 - 8))/(4) `

X = `(-6 +- sqrt (28)) / 4`

The solutions are as follows:


X =`(-6 + sqrt (28)) / 4`           and X =  `(-6 - sqrt (28)) / 4`

sine rule formula

In the trigonometry, Sine rule formula (also called as sine formula or law of sines or sines law or sine rule) is the equation that can be used for relating the lengths of the sides of the arbitrary triangle to sines of the corresponding angle. According to the sine rule formula,

`a/sin A = b/sin B =c/sin C`

The sine rule formula in the trigonometry defines that the ratio of Sine of the angle with respect to its included sign and the vice versa depending on number of the properties in an obtuse triangle. If the obtuse triangle has two sides with given included angle as SSA(Side-Side-Angle) or ASA(Angle-Side-Angle) or AAS(Angle-Angle-Side). Then the Sine Rule Formula can be used to find the Angle of one of the sides.

a, b, and c are the lengths of sides of a given triangle, and A, B, and C are the opposite angles. Sometimes the sine rule formula is also stated as the reciprocal of this equation:

`sin A/a = sin B/b =sin C/c`

The sine rule formula is used to compute the remaining sides of the triangle when the two angles and one side is known. The technique is known as triangulation. It is used when two sides and one of the non-enclosed angle is known. In such cases, the sine rule formula gives two possible values for an enclosed angle.

Proof of the Sine rule formula


From the above triangle AXC,

`sin A=h/b`

`bsin A=h`

 From triangle XBC,

`"sin B `

`asin B=h`

Equating both the Equations we have

`h=bsin A=asin B`

So,

`b/sin B=a/sin A`

 A perpendicular from A to BC, we can show that

`b/sin B=c/sin C`

Hence we have the Sine Rule:

 `a/sin A = b/sin B =c/sin C`

The area of a triangle :

The Area of any triangle is `1/2 ab sinC` using the sine rule formula.

The Sine rule formula can be used if we don't know the height of a triangle (since we have to know the height for

`1/2 (base) * (height)` .


Examples using the sine rule formula


Q 1: Given side a = 10, side c = 14, and angle C = 30°

Sol: Using the Sine rule formula , we conclude that

`sin A / 20 = sin 30 /14`

`sin A = 1/2 * 20/14`

`A = sin ^-1 (1/7)`

A = 0.14°

2)  Applying the sine rule formula :

`17/sin 62 = 13/sin theta`

`17 sin theta = 13 sin 62`

`sin theta = (13 sin 62)/ 17`

`sin theta = (13 * 0.8829)/17`

`sin theta = 0.6752`

`theta = sin ^-1 (0.6752)`

? = 42.47°

Solving Online Rectangle Properties

A rectangle is any quadrilateral with four right angles. The term oblong is generally used to denote to a non-square rectangle. A rectangle with vertices WXYZ would be denoted as WXYZ. A so-called crossed rectangle is a crossed quadrilateral which consists of two opposite sides of a rectangle along with the two diagonal.





(source: Wikipedia)

Properties of Rectangle - Solve Online Rectangle Properties

Symmetric Properties:

cyclic               - The four corners touch the same circle
Equiangular  - The corner angles are equal (i.e.) 90 degrees
Isogonal        - All corners are present within the same symmetry orbit

Possess reflectional symmetry  and rotational symmetry

Rectangle - Rhombus Duality:

Dual polygon of a rectangle = A rhombus



Formula:

Area = l `xx ` b  sq.units
Perimeter = 2 ( l + b ) units
Length of diagonal = `sqrt(l^2+b^2)`
when l = b, the rectangle becomes square

Example Problems on Solve Online Rectangle Properties

Find the perimeter of the rectangle of sides 4m and 1m

Solution:

Step 1: Let, l = 4m, b = 1m

Step 2: Perimeter of the rectangle = 2 (l + b)

Step 3: Perimeter of the rectangle = 2 (4+ 1)

= 2 (5)

= 10m

Result: Therefore, Perimeter of the rectangle = 6m

2.  Find the area of the rectangle of sides 4m and 1m

Solution:

Step 1: Let, l = 4m, b = 1m

Step 2: Area of the rectangle = l `xx` b

Step 3: Area of the rectangle = 4 `xx` 1

= 4 m2

Result: Therefore, Area of the rectangle = 4 m2

3.  Find the length of a diagonal of  the rectangle of sides 1m and 4m

Solution:

Step 1: Let, l = 4m, b = 1m

Step 2: Length of a diagonal of the rectangle =   `sqrt(l^2+b^2)`

Step 3: Length of a diagonal of the rectangle = `sqrt(4^2+1^2)`

= 16.03 m2

Result: Therefore, Length of a diagonal of the rectangle = 16.03 m2

Practice problems on Solve online Rectangle Properties:

Find the perimeter of the rectangle of sides 2cm and 3 cm
Find the area of the rectangle of sides 2cm and 3cm
Find the length of a diagonal of  the rectangle of sides 2cm and 3cm

Solutions for practice problems on Solve online Rectangle Properties:

10 cm
6 cm 2
3.6cm

Power of a Point Theorem

In math, power of point theorem is a theorem related to circle. Power of point theorem tells the relationship between the intersecting lines of a circle. Base on the intersecting lines, we have three possibilities for power of point theorem. This article gives a clear explanation of power of point theorem with some example problems.

Explanation to Power of Point Theorem;

Three possibilities for power of point theorem:
Case: 1



In the above figure, there are two intersecting lines intersect the circle insides. Then the power of point theorem is AE . CE = BE. DE

Case 2:



In the above figure, one of the line is tangent to the given circle. Then the power of point theorem is, AB2 = BC . BD

Case 3:



In the above figure, there are two lines intersect outside of the circle. Then the power of point theorem is CB . CA = CD . CE

Special case of power of point theorem:



In the above figure, there are two tangent lines. Then the power of point theorem is PA = PC

Example Problems to Power of Point Theorem:

Example: 1

Determine the unknown value of the following figure using power of point theorem.



Solution:

Given:

CB = 2

AB = 4

CD = x

DE = 1

Step 1:

Here two lines are intersecting at outside. So, the power of point theorem is, CB . CA = CD . CE

Step 2:

CB . CA = CB . (CB + CA)

= 2 . (2 + 4)

=2 . 6

= 12

Step 2:

CD . CE = x . (CD + DE)

= x . (x + 1)

= x2 + x

Step 3:

CB . CA = CD . CE

12 = x2 + x

x2 + x - 12 = 0

(x + 4)(x - 3) = 0

x = -4, 3 (Discard the negative answer)

x = 3

Answer: 1

Example: 2

Determine the unknown value of the following figure using power of point theorem.



Solution:

Given:

EA = x

ED = 2

EC = 5

EB = 5

Step 1:

Here two lines are intersecting at inside. So, the power of point theorem is,AE . CE = BE. DE

Step 2:

x . 5 = 5 . 2

5x = 10

x = `10/5`

x = 2

Answer: x = 2

Example Problems to Power of Point Theorem:

Problem: 1

Determine the unknown value of the following figure using power of point theorem.



Answer: 4

Problem: 2

Determine the unknown value of the following figure using power of point theorem.



Answer: 6

Sine Calculator

In mathematics sine is based on trigonometry. The trigonometry is the branch of mathematics it   deals with dealings between the angles of a triangle and sides. The sin determines the sides of the quadrilateral and it provisions the length of the residual diagonal. It is nothing but reformulation of Ptolemy’s theorem. Let u see sin calculator.

Sin Basic Addition Proof:

Let us see sin addition calculator. It one of the basic operations Here we can see sin addition formula proof.


Ptolemy’s theorem:



Here    BC = 1 AB= cos (α)

AC = sin (α) BD= cos (β)

DC = sin (β) AD = sin (α + β)

Therefore sin (α + β) = sin (α) cos (β) + cos (α) sin (β)

Sin addition formula: sin (a + b) = sin (a) cos (b) + cos (a) sin (b)



Sin the area of a triangle is half the creation of the elevation and the base, it follows that the area of a triangle is half the creation of any of the two side times sin of the included angle. For the three triangles in the diagram we have,

2Ac= c (d cosb) sin a,

2Ad= d (c cos a) sin b,

2A= cd cos (a + b)

This after cancelling cd gives the addition formula for sin.  Following examples are showing Sin values in calculator.


Sine Calculator:

Let us see find sin value in calculator. Following statements are showing briefly explanation.



Example 1:

Find sin 6 value in calculator:

Solution:

Using sin table sin 6 = .1045

Example 2:

Find sin 6 value in calculator:

Solution:

Using sin table sin 45 = .7071

Definition of Cos

Cos is one of the trigonometric ratios. The six basic trigonometric ratios are sin, cos, tan, sec, cosec and cot. Of these sin and cos are most familiarly used. Trigonometric ratios involve angles in their standard positions. They can be defined using a Right triangle as well as by a Circle.

Let us see the Definitions of Cos

Definition 1: using a right triangle



The sides of a right triangle are

1. Hypotenuse : The side opposite to the right angle (`90^o` ), in this case side `BC`
2. Opposite side: The side opposite to the angle of consideration (angle `theta` ), in this case side `AB`
3. Adjacent side: The side adjacent to the angle of consideration(angle `theta` ), here it is side `AC`

Definition of cos : Cosine is defined as the ratio of adjacent side to the hypotenuse

So, `cos(theta) = (adjacent side) / (hypoten use)`

`= (AC)/(BC)`

Here cos is defined for angles  `0^o< theta < 90^o`

Note: here ratio does not depend on the lengths of a particular triangle, it is same for all the right triangles which contain the  angle `theta` . Because all such right triangles will be simillar.

Definition 2: Using  Circle

Consider a circle with center  ‘`O` ’ at origin and radius ‘`r` ’ in the rectangular cartesian system. Let `theta` be any angle in standard position such that its terminal ray intersects the circle in point P(x,y).



From the figure

We have `OP = r`

And `x^2+y^2 = r^2`

Now cosine is defined as   `cos(theta)=x/r`

Here cos is defined for angles `0^o < theta < 360^o`

Examples for Definition of Cos :

For  a triangle  with sides AB =3, BC=5 ,AC=4 Right angles at ‘`A` ’


Calculate  `CosB` and `CosC`

Solution:

For calculating `cosB` :



Adjacent side is AB =3 , opposite side is AC =4, Hypotenuse is BC= 5

So `cos B = (adjacent side)/( hypoten use)`

`=(AB)/(BC)`

` = 3/5`

For calculating `Cos C` :



Adjacent side is AC =4, Opposite side is AB =3, Hypotenuse BC =5

So, `cos C = (adjacent side)/ (hypoten use)`

` = (AC)/( BC)`

` = 4/5`

Learning Natural Tangent Tables

The branch of math that deals with the relations of the sides and angles of triangles is known as trigonometry. Which the methods of reducing from a certain given parts to other required parts, and also in the general relations which exist between the trigonometrical functions of arcs or angles. The trigonometry ratio table is used to find the values of the trigonometry functions. Here we are going to see about learning of natural tangent tables:

The natural tangent of an angle is the ratio, the length of the opposite side to the length of the adjacent side. In our case

Tan A = `(opposite side) / (adjacent side)`

The natural tangent angle table will be very useful for finding the value of tangent angles. Learning of natural tangent tables will be very helpful when we are using the problems.


Learning Natural Tangent Tables:

Angle   Tangent(A)

Angle    Tangent (A)

0         0.000 1         0.017 2         0.035 3         0.052 4         0.070 5         0.087 6         0.105 7         0.123 8         0.141 9         0.158 10        0.176 11        0.194 12        0.213 13        0.231 14        0.249 15        0.268 16        0.287 17        0.306 18        0.325 19        0.344 20        0.364 21        0.384 22        0.404 23        0.424 24        0.445 25        0.466 26        0.488 27        0.510 28        0.532 29        0.554 30        0.577 31        0.601 32        0.625 33        0.649 34        0.675 35        0.700 36        0.727 37        0.754 38        0.781 39        0.810 40        0.839 41        0.869 42        0.900 43        0.933 44        0.966 45        1.000

45        1.000 46        1.036 47        1.072 48        1.111 49        1.150 50        1.192 51        1.235 52        1.280 53        1.327 54        1.376 55        1.428 56        1.483 57        1.540 58        1.600 59        1.664 60        1.732 61        1.804 62        1.881 63        1.963 64        2.050 65        2.145 66        2.246 67        2.356 68        2.475 69        2.605 70        2.747 71        2.904 72        3.078 73        3.271 74        3.487 75        3.732 76        4.011 77        4.331 78        4.705 79        5.145 80        5.671 81        6.314 82        7.115 83        8.144 84        9.514 85        11.430 86        14.301 87        19.081 88        28.636 89

Examples for Learning Natural Tangent Tables:

Find the values of the given natural tangent angles using natural tangent table.

1) tangent 1   =      0.017

2) tangent 48   =   1.1114

3) tangent 9   =     0.158

4) tangent 56  =     1.483

5) tangent 16   =     0.287

6) tangent 64   =     2.050

7) tangent 25    =    0.466

8) tangent 78    =    4.705

9) tangent 44   =    0.966

10) tangent 88   =    28.636