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