Triangle Geometry Cosine

Look at this geometry figure:



In all these right angled triangles of different sizes, <A has the same measure. Since <B = 90o, we have <C = 90o - <A, in each of these triangles. So, all these triangles have equal angles and hence are similar. Thus even though the lengths of the sides BC, CA, AB are different in these, the numbers `(BC)/(AC)` and `(AB)/(AC)` are the same for all these triangles.

In all these triangles, AC is the hypotenuse. The side BC is the opposite side of <A. The third side AB is called the adjacent side of<A.    In a right angled triangle with <A as one of its angles, the number obtained by dividing the adjacent side of  <A by the hypotenuse is called the cosine of <A and its written cos A. Thus

cos A = `(adjacent- side- of- angleA) / ( hypotenuse )`

Note that cos A is a number indicating the size of <A alone and not depend on the size of the right angled triangle containing <A.

Examples for Triangle Geometry Cosine:

Example 1:

Determine the cosine of angle A or cos A; If <A =45o in the right angled triangle ABC, as in the figure below



then, as seen earlier, we have

`(BC)/(AC)` = `(AB )/(AC)` =1/`sqrt(2)`

which in our new terminology can be written

cos 45o = 1/`sqrt(2)`

Example 2:

Determine the cos30o, if <A =30o in the right angled triangle ABC, as in the figure below



Solution:

By definition, we know that   cos A = `(AB)/(AC)`

Therefore,   cos 30 = `sqrt(3)` / 2


Example 3:

Measure the cos C, if <C =60o in the right angled triangle ABC, as in the figure below





Solution:

Given:

`Delta` ABC , AB = `sqrt(3)` , BC = 1 and AC = 2.

To find the triangle geometry cosine C:

By definition, we know that,  cos C = `(BC)/(AC)`

That is                                     cos 60o = `1/2`

This is the required triangle geometry cosine values.

Example 4:

Measure the cosine of angle A or cos A; If <A =20o, AB = 4.7 and AC =5 in the right angled triangle ABC, as in the figure below



Solution:

Given:

In `Delta` ABC, <A = 20o , AB = 4.7 and AC =5.

To find the triangle geometry cosine of angle A:

We have found by actual measurement that in such a triangle as `Delta` ABC

`(AB)/(AC)` = `(4.7) /(5)`  = 0.94

This means in other words

cos 20o = 0.94

Quadratic Functions in the Real World

Quadratic functions have their general form as the one shown below:

f(x) = a.x2 + b.x + c


Here we can see how quadratic functions are used in in the real world.
In the coordinate Geometry when this equation is plotted we get a parabola. And the shape, size and other dimensions are decided by the coefficients of the variables in the quadratic function i.e on a, b and c. And the solution of the equation in the general form is given by:

x =  ( -b ± √(b2 - 4ac)) / (2a)

where discriminant  d = (b2 - 4ac)

A quadratic equation with real coefficients can have either one or two distinct real roots, or two distinct complex roots. In this case the discriminant determines the number and nature of the roots
Quadratic Functions in the Real World- Physics and Maths

Quadratic Functions in Physics:

Projectile: A projectile is an object upon which the only force acting is gravity. A stone when thrown upwards with a velocity returns to ground after sometime. When the path of the projectile is traced it's a parabola. So when something is thrown up and allowed to fall down freely later we get the projective path as a parabola.
Basing on the properties of the parabola and the path traversed further observations are done for  studying the body motion in physics.

The volcanic eruptions also generally follow the projectile motion since the hot fluid masses or the lava is pumped out with great velocities. So Quadratic equations also have a dominant role in estimating the safe zones in case of an eruptions and for further assistance and analysis.

Quadratic Functions in Maths:

Non-normal distributions in probability cover good applications of parabolas when plotted. The distribution plots mostly have their distribution curves as parabolas.
Other real life applications would be parabolic mirror, shape of a spillway for a dam and sound reflector.

Another application of a quadratic equation used in higher math and engineering, where a second-order differential equation is solved for a spring-damper system. This sounds scary but actually has real-world application. This example also shows how the "imaginary" number "i" is used in a real-world application. The fact that the exponent is ^2 while the inputs are ^1 indicates that it would solve perimeter vs area (fencing) problems, or bill of materials issues for tanks and containment etc.

Quadratic Functions in the Real World -networking

Quadratic Functions in Computers Networking

Quadratic equations are used considerably in computer networking. They're used for all sorts of error checking, as well as encryption. They're also used in the mathematics of Quality of Service (Qos), the concepts of understanding how to dynamically allocate and share bandwidth.

In this way we can have lot many Quadratic Functions in the real World.

Siding Square Calculator

Let us see about siding square calculator.  The square feet are one of the measurements of siding the square. The measurement tool of square feet is denoted as sq ft. The square foot is the plural term of the square feet. The length of the siding square is calculated for the 2 dimensional objects.

Formula for the Siding Square:

Let us see the siding square formula’s which is used by the calculator. The formulas are described as below.

1. Siding square of the rectangle     = b * h.
2. Siding square of the square         = a * a.

These are all the formulas which are helped to find the siding square.

Examples of Siding Square:

Let us see some examples of siding square which are used by the calculator.

Example 1:

Find the siding square for the rectangle with the help of calculator, where the base has 8ft and the height is 7ft.

Solution:

Step 1: First, choose the siding square object.



Step 2: Enter the base length in the calculator.



Step 3: Enter the height of the rectangle.



Step 4: Choose the calculate option in the calculator. The calculation is done by the formula of b * h.



Step 5: Choose the exit button after getting the siding square.

These are procedures used to calculate the siding square of the rectangle.


Example 2:

Find the siding square for the square with the help of calculator, where the side’s lengths are 12ft.

Solution:

Step 1: First, choose the siding square object.



Step 2: Enter the side length of square in the calculator.





Step 3: Choose the calculate button in the calculator. The calculation is done by the formula of a * a.



Step 4: Choose the exit button after getting the siding square.

These are procedures used to calculate the siding square of the square.

Histogram with Uniform Width

Let a function f be continuous for every value of x in [a, b]. it means that if x0 ε [a, b], then given ε > 0, there exists δ > 0, such that

│f(x) –f(x0)│< ε whenever│ x – x0│< δ

The number δ will depend upon x0 as well as ε and so we may write it symbolically as δ (ε, x0). For some functions it may happen that given ε > 0 the same δ serves for all x0 ε [a, b] in the condition of condition of continuity. Such functions are called uniformly continuous on [a, b]

Definition for Uniform Continuity

A function f defined on an interval I is said to be uniformly continuous on I if given ε > 0, there exists a δ > 0 such that

│f(x) – f(y) │< wherever │x - y│< δ

It should be noted carefully that uniform continuity is a property associated with an interval and not with a single point. The concept of continuity is local in character where as the concept of uniform is global in character.

Note:- If f uniformly continuous on an interval I, then it is continuous on I.

Note:- A function which is continuous in a closed and bounded interval I = [a, b] is uniformly continuous in [a, b]
 
Examples on Uniform Continuity

Give an example to show that a function continuous in an open interval may fail to be uniformly continuous in the interval.

Solution

Consider the function f defined on the open interval ] 0, 1[ as follows:

We have, x is continuous in ] 0, 1[ and x ≠ 0 in ] 0, 1[

→ 1/x is continuous in ] 0, 1[

Now, for any δ > 0, we can find m `in`  N such that

1/n < δ for every n > m

Let x1 1/2m and x2 = 1/m so that x1, x2 `in` ] 0, 1[

│x1 – x2│= │1/2m – 1/m│= 1/2m < 1/2δ < δ

But │f(x1) – f(x2)│= │2m – m│= m

Which cannot be less than each ε >0

Example2

Show that the function f denoted by f(x) = x3,is uniformly continuous in [–2, 2]

Solution

Let x1, x2 `in`  [–2, 2].

We have │f(x2) – f(x1)│=│x23 – x13│

= │(x2 – x1) (x23 + x13 + x1x2│

<  │x2 – x1│{│x2│3+│x1│2+│x1││x2│}

< 12│x2 – x1│

... │f(x2) – f(x1)│< ε wherever│x2 – x1│< ε/12

Thus given ε > 0, there exists δ = ε/12 such that

│f(x2) – f(x1)│< ε whenever│x2 – x1│< δ `AA` x1, x2 `in`  [–2, 2]

Hence f(x) is uniformly continuous in [–2, 2]

Maximum Area of a Rectangle

We know that a rectangle is a form of a quadrilateral whose opposites sides are equal. The longer sides are called its length and their shorter sides are called width. The area of the rectangle is given by length * width.

To obtain a maximum area then length = width. So if the perimeter of the rectangle is given then we need to take the value of both length and width to be equal.

Now let us see few problems of this kind.

Example Problem on Maximum Area of a Rectangle:

Ex 1: Find the maximum area of a rectangle whose perimeter is 144cm.

Soln: Given: Perimeter = 144

`implies ` 2(L + W) = 144

`implies` L + W = 144/2 = 72

`implies` L + W = 72

`implies` L = W =36

Therefore, the maximum area = 36 `xx` 36 = 1296cm2


 
Ex 2: If the maximum area of a rectangle is 2916cm2. Find the length and width of the rectangle.

Soln: Given: Area = 2916

Since the maximum area of rectangle is obtained by having length equal to the width,

Area = l * w = 2916

` implies` l * l = 2916    [since l = w]

`implies`         l2 = 2916

` implies`        l = `sqrt[2916]` = 54

Therefore the dimension is 54 * 54 to get the Maximum area.

More Example Problem on Maximum Area of a Rectangle:

Ex 3: Perimeter of a rectangular fence is 252cm. Find the maximum area the fence can cover.

Soln: Given: Perimeter = 252

2( L + W) = 252

`implies` L + W = `252/2` = 126

`implies` L = W = `126/2` = 63

The maximum Area = 63 * 63 = 3969 m2

Ex 4: Let the length of the rectangle be 91cm. Its area is 8281 cm2. Can this length form a maximum area?

Soln: Given: Area = 8281 cm2 and L = 91

`implies` L `xx` W = 8281

`implies` W =` 8281/91` = 91

Here since w = 91 = L, 8281cm2 is the maximum area that can be formed by the length of 91cm.

Linear Input Output Function

Linear input output functions is nothing but linear equations in terms of  functions. The expression deals with linear equation of a particular function. Function notation will look like f(x), p(x) form. Linear function are expressed as f(x) = ax+b. This is how the linear function be displayed with the input value.  We need to find the output linear function.

Example:

f(3)  = 3x+4 it is the linear equation in the form of ax+b. The input function is given as 3, we need to find the output linear function.

Problems in Linear Input Output Function:

Example 1:

Simplifying the linear input output function of f(x) = 3x+12. The input function is f(4)

Solution:

In the given equation substitute the input value,

f(x) = 3x+12 the linear input function is 4

f(4) = 3 * 4 + 12. here multiply the 3 * 4 is 12 and add.

f(4) = 12+12

f(4) = 24

The linear output function is 24.

 Example 2:

Simplifying the linear input output function of f(x) = 4x+20. The input function is f(5)

Solution:

In the given equation substitute the input value,

f(x) = 4x+20 the linear input function is 5

f(5) = 4*5+20. here multiply the 4*5 is 20 and add.

f(5) = 20+20

f(5) = 40

The linear output function is 40.

Problems in Linear Input Output Function:

Example 4:

Simplifying the linear input output function of f(x) = 13x+22. The input function is f(4)

Solution:

In the given equation substitute the input value,

f(x) = 13x+22 the linear input function is 4

f(4) = 13*4+22. here multiply the 13*4 is 52and add.

f(4) = 52+22

f(4) = 74

The linear output function is 74.

Example 4:

Simplifying the linear input output function of f(x) = 4x+5. The input function is f(5)

Solution:

In the given equation substitute the input value,

f(x) = 4x+5 the linear input function is 5

f(5) = 4*5+5. here multiply the 4*5 is 20 and add.

f(5) = 20+5

f(5) = 25

The linear output function is 25.

Radius is Half the Diameter

In geometry, a diameter of a circle is any straight line segment that passes through the center of the circle and whose endpoints are on the circle. The diameters are the longest chords of the circle. The word "diameter" derives from Greek (diametros), "diagonal of a circle", from (dia-), "across, through" + (metron), "a measure. (Source wikipedia)



Radius is defined as the half of the diameter .we can write

Radius = half of the diameter or `(diameter) / 2`

Here we are going to study about how to solve the diameter problems and its example.

Radius is Half the Diameter – Example Problems:

Example: 1

Calculate the area of the circle with diameter of the circle is 39 meter.

Solution:

We know the formula for calculate the area of the circle = `pi ` r2

In this problem diameter is given so we have to find the radius we know the radius is half of the diameter so we have to divided by two

Radius = `(diameter) / (2)`

r = `39/2`

r = 19.5 meter

Now we find the area of the circle

Area = 3.14 * 19.52

Area = 3.14 *19.5 * 19.5

Simplify the above we get

Area = 854.86 meter square

Therefore the area of the circle is 1193.98 meter2


Radius is Half the Diameter – Example: 2

Find the surface area of the sphere with diameter is 23 cm

Solution:

We know that formula for finding surface are a of the sphere = 4 `pi` r2

In this problem diameter is given so we have to find the radius we know the radius is half of the diameter so we have to divided by two

Radius = `23 /2`

r = 11.5 meter

Now we find the surface area of the sphere

Area = 4 *3.14*11.52

Area = 4 *3.14*11.5*11.5

Simplify the above we get

Area = 1161.06 meter square

Therefore the surface area of the sphere = 1161.06 meter2