Category: Maths Notes PPT

A sample statistic refers to quantity from the sample of the given population. A sample is a group of elements that are chosen from the population. The features which we use to describe the population are called the parameters and the properties of the sample data are known as statistics. Population and sample both are the important part of statistics. A sample statistic is a piece of information that we collect from a fraction of a population. Here we will study about sampling statistics methods, hypothesized mean, mean standard deviation and distribution of means.

What is a Sample in Statistics?

A sample statistic is a numerical descriptive measure of a sample data points. A statistic is generally derived from measurements of the individual data from the sample. The statistics are a characteristic of a sample data distribution such as mean, median, mode, standard deviation and proportions. A sample statistic can be used to measure any characteristic of the sample.

Hypothesized Mean

Hypothesis testing is an essential procedure in statistics. A hypothesis test used to evaluate two mutually exclusive statements about a population that determine which statement is best and also supported by the sample data.

The process of hypothesis testing involves setting up two competing hypotheses, first is null hypothesis and second one is alternate hypothesis.

The techniques for hypothesis testing depend on

(i) the type of outcome variable being analyzed (continuous, dichotomous, discrete)

(ii) the number of comparison groups in the investigation

(iii) whether the comparison groups are independent

Estimating the Mean

Following are the steps for estimating the mean :

Step 1. First we have to add a new column to the table writing down the midpoint (middle value) of each group.

Step 2.  Multiply each midpoint value by the frequency of that group and then add the results in a new column.

Step 3. Add the values in the midpoint × frequency column.

Step 4. Finally, divide that value by the total frequency to get the estimate of the mean.

Sample Standard Deviation

The sample standard deviation formula is:

s = [sqrt{frac{sum (X – bar{X})^{2}}{n-1}}]

Sample standard deviation formula

where,

s = sample standard deviation

[sum] = sum

[bar{X}] = sample mean

n = number of scores in the sample.

Sampling Distribution

A sampling distribution is similar to a probability  distribution of a statistic that we choose from random samples of a given population. It is also known as a finite-sample distribution, it represents the distribution of frequencies for how to spread apart various outcomes for a specific population.

The sampling distribution depends on multiple factors such as statistics, sample size, sampling process, and the overall population. It is used to help calculate statistics such as means, ranges, variances and standard deviations for the given sample.

Sample Mean

The sample mean refers to the average value found in a sample. A sample is just a small part of a whole data. For example, if we work for a polling company and want to know how much people pay for food a year, you aren’t going to want to poll over 300 million people. Instead of that, we take a fraction of that 300 million (perhaps a thousand people) that fraction is called a sample. In other words, mean refers to “average.” So in this example, the sample mean will be the average amount therefore those thousand people will have to pay for food a year.

The sample mean is useful when we have to estimate what the whole population is doing, without surveying everyone. Suppose sample mean for the food example was ＄2400 per year. The odds that we will get is a very similar figure if we surveyed all 300 million people. So the sample mean is a way to save a lot of time as well as money.

The Sample Mean Formula 

The sample mean formula is: [bar{X}] = [frac{sum x_{i}}{n}]

Here

• [bar{X}] just stands for the “sample mean”

• [sum] is summation notation

• x[_{i}] “all of the x-values”

• n is number of items in the sample mean

Mean and Standard Deviation

The mean refers to average or the most common value in a collection of numbers. There are multiple ways to calculate the mean. There are the two most popular methods i.e Arithmetic mean and geometric mean.

A standard deviation is the measurement of the distribution of a dataset which is related to its mean and it is calculated by the square root of the variance. It is calculated as the square root of variance by determining each data point’s deviation which is relative to the mean. If the data points are further from the mean, then there is a chance of higher deviation within the data set. Therefore, the more spread out the data, the higher is the standard deviation.

The Formula for Standard Deviation is Given Below:

Standard deviation = s = [sqrt{frac{sum_{i=1}^{n} (X_{i} – bar{X})^{2}}{(n-1)}}]

Where 

X[_{i}] = It is the of the i[^{th}] point in the data set

[bar{X}] = It is the mean value of the data set

X = It is the  number of data points in the data set

Probability Sample

Probability sampling is a sampling technique that is used by researchers to choose samples from a larger population using a method that is based on the theory of probability. For a participant to be considered as a probability sample, they must be selected using a random selection.

The most critical requirement of probability sampling is that everyone in the population is known and they have equal chance of getting selected. Suppose, if we have a population of 100 people, and every person would have odds of 1 in 100 for getting selected. In this case probability sampling gives us the best chance to create a sample that is mainly representative of the population.

It uses statistical theory while selecting a small group of people (or sample) from an existing large population and then predicts all their responses that will match with the overall population.

Errors in Sampling

Sampling error often occurs when the sample we use in the study is not representative of the whole population. It often occurs, that’s why, researchers always calculate a margin of error during final results as a statistical practice. The margin error is the amount of error that is allowed for miscalculation while representing the difference between the sample and the actual population. We can control and eliminate these sampling by creating a sample design, having a large enough sample to reflect the entire population, or using an online sample or survey audience
to collect responses. 

Before knowing what a second-order derivative is, let us first know what a derivative meAns: Basically, a derivative provides you with the slope of a function at any point. The derivative of the first derivative of a function is known as the second-order derivative. The slope of the tangent at a given location, or the instantaneous rate of change of a function at that position, is determined by the first-order derivative at that point. Second-Order Derivative offers us an understanding of the shape of a function’s graph. The second derivative of the function f(x) is commonly abbreviated as f” (x). If y = f, it is sometimes expressed as D²y or y² or y” (x).

Let’s say y = f. (x)

dy/dx = f’ then (x)

If f'(x) is differentiable, we can differentiate it with respect to ‘x’ once more. The left-hand side thus becomes d/dx(dy/dx), often known as the second-order derivative of y w.r.t x.

Now, what is a second-order derivative? A second-order derivative is a derivative of the derivative of a function. It is drawn from the first-order derivative.  So we first find the derivative of a function and then draw out the derivative of the first derivative. A first-order derivative can be written as f’(x) or dy/dx whereas the second-order derivative can be written as f’’(x) or d²y/dx² 

A second-order derivative can be used to determine the concavity and inflection points. 

Concavity

Concave Up: The second derivative of a function is said to be concave up or simply concave, at a point (c,f(c)) if the derivative  (d²f/dx²)x=c >0. In such a case, the points of the function neighboring c will lie above the straight line on the graph which will be tangent at the point (c, f(c)). Here is a figure to help you to understand better. 

uploaded soon)

Concave Down: Concave down or simply convex is said to be the function if the derivative (d 2 f/dx²)x=c at a point (c,f(c)). In such a case, the points of the function neighboring c will lie below the straight line on the graph which is tangent at the point (c,f(c)). Here is a figure to help you to understand better. 

uploaded soon)

Point of Inflection

The point of inflection can be described as a point on the graph of the function where the graph changes from either concave up to concave down or concave down to concave up. The sigh of the second-order derivative at this point is also changed from positive to negative or from negative to positive. The second-order derivative of the function is also considered 0 at this point. 

uploaded soon)

Second-Order Derivative Examples

Question 1) If f(x) = sin3x cos4x, find  f’’(x). Hence, show that,  f’’(π/2) = 25.

Solution 1) We have, 

f(x) =  sin3x cos4x or, f(x) = [frac{1}{2}] . 2sin3x cos4x = [frac{1}{2}](sin7x-sinx) 

Differentiating two times successively w.r.t. x we get, 

f’(x) = [frac{1}{2}] [cos7x . [frac{d}{dx}]7x-cosx] = [frac{1}{2}] [7cos7x-cosx]

And f’’(x) = [frac{1}{2}] [[7(-sin7x)frac{d}{dx}7x-(-sinx)]] = [frac{1}{2}] [-49sin7x+sinx]

Therefore,f’’(π/2) = [frac{1}{2}] [-49sin(7 . π/2)+sin π/2] = [frac{1}{2}] [-49 . (-1)+1]

[sin7 . π/2 = sin(7.π/2+0) = – cos0= -1]

                   = [frac{1}{2}] x 50 = 25(Proved)

Question 2) If y = [tan^{-1}] ([frac{x}{a}]), find y₂.

Solution 2) We have,  y = [tan^{-1}] ([frac{x}{a}])

Differentiating two times successively w.r.t. x we get,

y₁ = [frac{d}{dx}] ([tan^{-1}] ([frac{x}{a}])) = [frac{1}{1+x²/a²}] . [frac{d}{dx}]([frac{x}{a}]) = [frac{a²}{x²+a²}] . [frac{1}{a}] = [frac{a}{x²+a²}]

And, y₂ = [frac{d}{dx}] [frac{a}{x²+a²}] = a . [frac{d}{dx}] (x²+a²)^-1 = a . (-1)(x²+a²)^-2 .  [frac{d}{dx}] (x²+a²)  

             = [frac{-a}{ (x²+a²)²}] . 2x = [frac{-2ax}{ (x²+a²)²}] 

Question 3) If y = [e^{2x}] sin3x,find y’’. 

Solution 3) We have, y = [e^{2x}]sin3x 

Differentiating two times successively w.r.t. x we get,

y’ = [frac{d}{dx}]([e^{2x}]sin3x) = [e^{2x}] . [frac{d}{dx}]sin3x + sin3x .  [frac{d}{dx}] [e^{2x}]

Or,   

y’ = [e^{2x}] . (cos3x) . 3 + sin3x . [e^{2x}] . 2 = [e^{2x}] (3cos3x + 2sin3x)

And    

y’’ = [e^{2x}][frac{d}{dx}](3cos3x + 2sin3x) + (3cos3x + 2sin3x)[frac{d}{dx}] [e^{2x}]

        = [e^{2x}][3.(-sin3x) . 3 + 2(cos3x) . 3] + (3cos3x + 2sin3x) . [e^{2x}] . 2  

        = [e^{2x}](-9sin3x + 6cos3x + 6cos3x + 4sin3x) =  [e^{2x}](12cos3x – 5sin3x)

Question 4) If y = acos(log x) + bsin(log x), show that,

                     x²[frac{d²y}{dx²}] + x [frac{dy}{dx}] + y = 0

Solution 4) We have, y = a cos(log x) + b sin(log x)

Differentiating both sides of (1) w.r.t. x we get,

[frac{dy}{dx}] = – a sin(log x) . [frac{1}{x}] + b cos(log x) . [frac{1}{x}]

Or, 

x[frac{dy}{dx}] = -a sin (log x) + b cos(log x)

Differentiating both sides of (2) w.r.t. x we get,

x . [frac{d²y}{dx²}] +  [frac{dy}{dx}] . 1 = – a cos(log x) . [frac{1}{x}] – b sin(log x) . [frac{1}{x}]

Or,     

x²[frac{d²y}{dx²}] + x[frac{dy}{dx}] = -[a cos(log x) + b sin(log x)]

 Or,  

x²[frac{d²y}{dx²}] + x[frac{dy}{dx}] = -y[using(1)]     

Or, 

x²[frac{d²y}{dx²}] + x[frac{dy}{dx}] + y = 0 (Proved)   

Question 5) If y = [frac{1}{1+x+x²+x³}], then find the values of

[frac{dy}{dx}]x = 0 and [frac{d²y}{dx²}]x = 0

Solution 5) We have, y = [frac{1}{1+x+x²+x³}] 

Or,  

y =   [frac{x-1}{(x-1)(x³+x²+x+1)}] [assuming x ≠ 1]    

    = [frac{x-1}{(x⁴-1)}]       

Differentiating two times successively w.r.t. x we get,

[frac{dy}{dx}] = [frac{(x⁴-1).1-(x-1).4x³}{(x⁴-1)²}] = [frac{(-3x⁴+4x³-1)}{(x⁴-1)²}]…..(1)

And,  

[frac{d²y}{dx²}] = [frac{(x⁴-1)²(-12x³+12x²)-(-3x⁴+4x³-1)2(x⁴-1).4x³}{(x⁴-1)⁴}]…..(2)

Putting x = 0 in (1) and (2) we get, 

[frac{dy}{dx}] x = 0 = [frac{-1}{(-1)²}] = 1 and [frac{d²y}{dx²}] x = 0 = [frac{(-1)².0 – 0}{(-1)⁴}] = 0

Second-Order Derivatives of a Parametric Function

We utilize the chain rule twice to determine the second derivative of the function in parametric form. To determine the second derivative, first, find the first derivative’s derivative with respect to t, then divide by the derivative of x with respect to t. If x = x(t) and y = y(t), then the second-order parametric form is:

[frac{dy}{dx}] = [frac{(dy/dt)}{(dx/dt)}] is the first derivative.

[frac{d^{2}y}{dx^{2}}] = [frac{d}{dx (dy/dx)}] is the second derivative.

[frac{(dy/dx)}{ (dx/dt)}] =[frac{ d}{dt} frac{ (dy/dx)}{ (dx/dt)}]

Note: The formula [frac{d^{2}y}{dx^{2}}] = [frac{(d^{2}y/dt^{2})}{(d^{2}x/dt^{2})}]  is completely incorrect.

The local maximum or lowest inflection point values are determined by a function’s second derivative. These can be recognised using the following criteria:

• The function f(x) has a local maximum at x if f”(x) < 0.

• The function f(x) has a local minimum at x if f”(x) > 0.

• If f”(x) = 0, it is impossible to draw any conclusions about the point x.

You can solve the following second-order differential equation:

P(x)dy/dx + Q(x)y = f d²y/dx² + P(x)dy/dx + Q(x)y = f (x)

P(x), Q(x), and f(x) are functions of x, and they are calculated using:

Inconclusive If f(x) is a polynomial, exponential, sine, cosine, or a linear mix of these, it will only work.

Parameter variation, which is a little messier but works on a broader range of functions.

However, let’s start with the scenario where f(x) = 0 (which makes it “homogeneous”):

Allow f(x) to be a differentiable function in a convenient interval. 

The graph of f(x) can then be classified as:

At the point (c, f(c), the function is said to be Concave Up, or simply Concave, if the derivative (d²f/dx²)x=c>0. The points on the graph of the function in the vicinity of c lie above the straight line that is tangent at the point (c, f(c) in this example.

At the point (c, f(c), the function is said to be Concave Down, or simply Convex, if the derivative (d²f/dx²)x=c0. The points on the graph of the function in the vicinity of c are below the straight line that is tangent at the point (c, f(c) in this case.

The function rises at the point (c, f(c) if the derivative (df/dx)x=c>0. 

You also need to know how the derivative of a function changes when x varies. The function is considered to be Concave Up if the derivative acts as an increasing function, i.e. d/dx(df/dx)>0. The function is considered to be Concave Down if the derivative acts like a declining function, i.e. d/dx(df/dx)<0. As a result, the second derivative's value is critical in establishing the shape of the function's graph.

The second-order derivative at a given position (c, f(c) is computed if f'(x) = 0 at that point. It is a Local Minimum if f”(x) > 0 at that point, and it is a Local Maximum if f”(x) < 0 at that location. Higher-order derivatives or other methods of determination are required if f"(x) = 0.

Similar Figures are those figures which have the same shape, but the magnitude of their dimensions may or may not be equal. If the magnitude of their dimensions is also equal, then they are said to be congruent figures. All congruent figures are similar, but all similar figures are not congruent.

Some geometric forms are always identical in design. Imagine a circle, although the size of the object tends to change, the form stays the same. It may be assumed, thus, that all circles of varying radii are similar to each other. The figure shown below indicates the concentric circles whose radii are different, but all of them are identical. However, since their sizes are different, they are not congruent.

uploaded soon)

In-Depth Concept of Similar Figure

In general, the opinion to say that something is similar to something else is to mention that the two things share common characteristics. For example, you and you’re friend might be brainstorming on how to solve a particular problem, and you tell him the approach you would take. He then will tell you that your approach is similar to the plan he was thinking of. This means that your ideas on how to solve the problem are much the same but could have small differences as well.

In mathematics, saying that two figures are similar means that they share a common shape. They can be different sizes, but they must have the same shape.

 

Definition of Similar Figure

Two figures are explained to be similar if they are the same shape. In more mathematical language, two figures are similar if their corresponding angles are harmonious, and the ratios of the lengths of their equivalent sides are equal.

 

Similar Figures Real Life Examples

For example, in real life, the front wheels of a vehicle, the hands of a human, two teacups, etc. are representations of congruent figures or objects. All identical shape items have the same form, but the measurements are different. The ∼ sign is used to symbolize similarity.

 

Scale Ratio

For two similar figures, their corresponding dimensions are in a particular ratio. This ratio is called the scale ratio. Also, for similar figures, the corresponding angles are equal. 

Consider two similar triangles. They would be mathematically represented as:

uploaded soon)

∆PQR ~ ∆ SUV

Since they are similar, the following is true for them.

• PQ/SU = QR/UV = PR/SV = scale ratio

• ∠P = ∠S, ∠Q = ∠U, ∠R = ∠V (corresponding angles)

For congruent figures, their scale ratio is equal to one, because their dimensions are equal.

 

Similar Figures Area and Volume

If two figures are similar, then their corresponding sides are proportional or it can also be explained as when the ratio of their sides is equal.

 

For example; if we take the ratio of their surface areas, then it will be identical to the square of the ratio of sides. The ratio of the volume of two similar figures will be equal to the cube of the ratio of the length of sides.

 

Hence, based on the declarations mentioned above, the scale factors of area and volume can be represented as;

SFA = SF2

SFV = SF3

where SFA is the scale factor of surface area and SFV is the scale factor of volume

Example of Similar Figures:

Following are the examples of similar figures:

1. Pair of equilateral triangles

2. Pair of squares

3. Pair of circles

 

Solved Examples

Example 1: Prove that the ratio of the area of two similar figures is the square of the ratios of their sides using the below figures.

uploaded soon)

Solution:

Ratio of Sides = HE/DA = EF/AB

= 4/2 = 8/4

= 2 cm

Square of Ratio of Sides = 22 = 4

Area of Rectangle = l x b

Area (ABCD) = 4 x 2

= 8 cm2

Area (EFGH) = 8 x 4

= 32 cm2

Ratio of Areas = Area (EFGH) / Area (ABCD)

= 32 / 8

= 4cm2

Ratio of Areas = Square of Ratio of Sides = 4cm2

Hence, proved.

Example 2: The below two rectangles are similar. Find the ratio of their perimeters. Establish a relationship between the ratio of sides to the ratio of perimeters. 

Solution:

Ratio of Sides = HE/DA = EF/AB

= a/4 = 12/6  = 2

a = 4 x 2 = 8 cm 

Perimeter of Rectangle = 2 (l + b)

uploaded soon)

Perimeter (ABCD) = 2 (6 + 4)

= 2 (10)

= 20 cm

Perimeter (EFGH) = 2 (12 + 8)

= 2 (20) 

= 40 cm

Ratio of Perimeters = Perimeter (EFGH) / Perimeter (ABCD)

= 40 / 20

= 2cm

Ratio of Perimeters = Ratio of Sides = 2cm

This is the relationship between the ratio of perimeters to ratio of sides.

Example 3: The perimeters of two similar triangles is in the ratio 2 : 3. The sum of their areas is 169 cm2. Find the area of each triangle.

Solution:

We know that the perimeters of two similar triangles are in the ratio 2 : 3

Then,

The perimeter of the 1st  triangle =  2x

The perimeter of the 2nd  triangle = 3x

We know that the ratio of the areas is equal to the square of ratios of the sides, and the ratios of the sides are equal to the ratio of the perimeters. Hence, the ratio of the areas is equal to the square of ratios of the perimeters.

Area of 1st triangle: Area 2nd  triangle= (2x)2 : (3x)2

Area of 1st triangle: Area 2nd triangle=  4x2: 9x2

The sum of the areas is 169 cm2

Then, 4x2 + 9x2  =  169

13x2 = 169

x2 = 13

Therefore, the area of the triangles
are:

Area of 1st  triangle =  4(13) =  52 cm2

Area of 2nd triangle =  9(13) =  117 cm2

Conclusion

This is all about the concept of similar figures, its explanation with solved examples. Understand how similar figures are explained here and develop your concepts well. Focus on how the concept is used to solve problems.

Trigonometry is the branch of mathematics that studies the relationships between angles and sides of triangles or that deal with angles, lengths, and heights of triangles and relations between different parts of circles and other geometrical figures. The Concepts of trigonometry are very useful in practical life as well as finds application in the field of engineering, astronomy, Physics, and architectural design.

These are in total 6 ratios that we study in trigonometry used to tell us about the triangle and Sine and Cos are two of them. We will discuss in detail about the Sin Cos ratio, formula and other concepts.

Trigonometric Ratios

In mathematics six Trigonometric Ratios for the right angle triangle are defined i.e Sine, Cosecant, Tangent, Cosecant, Secant respectively. These Trigonometric Ratios are real functions which relate an angle of a right-angled triangle to ratios of two of its side lengths. Sin and Cos are basic Trigonometric functions that tell about the shape of a right triangle.There are six Trigonometric Ratios, Sine, Cosine, Tangent, Cosecant, Secant and Cotangent and are abbreviated as Sin, Cos, Tan, Csc, Sec, Cot. These are referred to as ratios because they can be expressed in terms of the sides of a right-angled triangle for a specific angle [theta].

• Adjacent: It is the side adjacent to the angle being taken for consideration.

• Opposite: It is the side opposite to angle being taken for consideration.

• Hypotenuse: It is the side opposite to the right angle of the triangle or the largest side of the triangle.
  

uploaded soon)

• Sin θ = [frac{Perpendicular}{Hypotenuse}] = [frac{Opposite}{Hypotenuse}]

• Cos θ =  [frac{Base}{Hypotenuse}] = [frac{Adjacent}{Hypotenuse}]

• Tan θ = [frac{Perpendicular}{Base}] = [frac{Opposite}{Adjacent}]

• Cosec θ =[frac{Hypotenuse}{Perpendicular}] = [frac{Hypotenuse}{Opposite}]

• Sec θ = [frac{Hypotenuse}{Base}] = [frac{Hypotenuse}{Adjacent}]

• Cot θ = [frac{Base}{Perpendicular}] = [frac{Adjacent}{Opposite}]
  

Basic Identities of Sine and Cos

If [A + B] = [90^{0}],that is A and B are complementary to each other then:

• [Sin (A) = Cos (B)]

• [Cos (A) = Sin (B)]

If [A + B] = [180^{0} ] then:

• [Sin (A) = Sin (B)]

• [Cos (A) = – Cos (B)]

     [ Cos^{2}(A) + Sin^{2}(A) = 1]

Double and Triple Angles

The Double or Triple Angle ratios can be converted to Single angle ratio of Sin and Cos using the below mentioned formulae:

• [Sin2A = 2 SinA.CosA ]

• [Cos2A = Cos^{2}A – Sin^{2}A = 2Cos^{2}– 1 = 1 − 2Sin^{2}A ]

• [Sin3A = 3 SinA – 4 Sin^{3}A ]

• [Cos3A = 4 Cos^{3}A – 3 CosA ]

• [Sin^{3}A = 4 Cos^{3}A SinA – 4 CosA Sin^{3}A ]

• [Cos4 A = Cos^{4}A – 6 Cos^{2}A Sin^{2}A + Sin^{4}A ]

• [Sin^{2} A = frac{(1 – Cos2A)}{2} ]

• [Cos^{2} A = frac{(1 + Cos2A)}{2} ]
  

Sum and Difference of Angles

The Sin and Cos ratio of sum and difference of two angles can be converted two product using below mentioned identities:

• Sin(A + B) = Sin(A) Cos(B) + Cos(A) Sin(B)

• Sin(A − B) = Sin(A) Cos(B) − Cos(A) Sin(B)

• Cos(A + B) = Cos(A) Cos(B) − Sin(A) Sin(B)

• Cos(A − B) = Cos(A) Cos(B) + Sin(A) Sin(B)

• Sin(A + B + C) = SinACosBCosC + CosASinBCosC + CosACosBSinC − SinASinBSinC

• Cos(A + B + C) = CosACosBCosC − SinASinBCosC − SinACosBSinC − SinACosBSinC − CosASinBSinC

• [SinA + SinB = 2Sin frac{(A + B)}{2}.Cosfrac{A – B}{2}]

• [SinA – SinB = 2Sinfrac{A – B}{2}.Cosfrac{A + B}{2}]

• [CosA + CosB = 2Cos frac{(A + B)}{2}.Cosfrac{A − B}{2}]

• [CosA + CosB = −2 Sin frac{(A + B)}{2}.Sinfrac{A − B}{2}]

In order to remember the trigonometric ratios values follow the given below steps:

We can the values for Sine ratios,i.e., 0, [ frac{1}{2}], [ frac{1}{sqrt{2}}], [ frac{sqrt{3}}{2}], and 1 for angles [0^{circ} ],[30^{circ} ],[45^{circ} ], [60^{circ} ]and [90^{circ} ] and the Cos ratio will follow the exact opposite pattern i.e. 0, [ frac{1}{2}], [ frac{1}{sqrt{2}}], [ frac{sqrt{3}}{2}], and 1 at[90^{circ} ],[60^{circ} ],[45^{circ} ],[30^{circ} ] and[0^{circ} ]. While the Tan will be the ratio of Sin and Cos ratio.

The value of Cosec, Sec and Cot is exactly the reciprocal of Sin, Cos and Tan.

Values of Trigonometric Ratios at Various Angles:

Trigonometry is a branch of mathematics and a sub-branch in algebra concerned with the measurement of specific functions of angles and their application to calculations. An example of trigonometry which is easy to understand is that of what architects use to calculate any particular distances.

Algebra and trigonometry are two major branches of mathematics. Algebra involves the study of math with specific formulas, rules, equations, and other variables. Trigonometry deals only with the triangles and their measurements.

The Six Main Functions of an Angle that are Commonly Used in Trigonometry are 

• sine (sin), 

• cosine (cos), 

• tangent (tan), 

• cotangent (cot), 

• secant (sec), and 

• cosecant (csc).

An easy and simple way to learn and understand Trigonometry is by studying all the basics of trigonometric angles and formulas by writing them down in a separate notebook which will be really useful to revise them before exams. Make sure you understand and study all the entire right-angle triangle concepts well so that you might compare any problems with a triangle before you try to solve them. The main thumb rule to score well in trigonometry is to learn your Pythagoras theorem with a whole heart. Keeping the Sine rule and Cosine rule at your fingertips will help you solve any type of problem in the examination. Finally, list down all the important identities and formulas of trigonometry in your mind and revision notes as well, and be thorough. Remember to learn how to use the trigonometry table in the necessary place.

Some applications of trigonometry class 10 notes chapter 9 are available with NCERT Solutions at . The notes are typically designed by the mathematics masters at the top-notch online education portal keeping in mind the updated pattern and guidelines by the CBSE board. These quick notes on CBSE class 10 maths will help you to significantly refine your trigonometric skills as well get to the core of the topic in no time.

In addition, you can also check other Maths learning resources such as previous year question papers, sample papers, activity quizzes etc. all for free download at .

Examples of Class 10 Chapter 9 – Some Applications of Trigonometry

Under this section, you will find the solved questions of chapter 9 – Some Applications of Trigonometry from Class 10 Maths textbook along with answer keys. These solutions are available for free PDF download from the given link at official. Let’s get started with the solved Maths Class 10 chapter 9 questions. 

Example: A villager is climbing a coconut tree using a 20 m long rope, which is tightly tied from the top of a vertical pole to the ground. Evaluate the height of the coconut tree, if the angle formed by the rope with the ground level is 30°. You can find the illustration below.

(Image to be added soon)

Solution: Given that length of the rope named AC = 20m

Angle formed is ∠ACB = 30°

Let the height AB of the coconut tree be h (meters)

With that, in right ▲ABC,

sin 30° = AB/AC

½= h/20 (since sin 30° = ½)

H = 20/2 = 10metres

Therefore, the height of the coconut tree is 10m

Example: The facilities department of a housing society plans to put two slides for the kids to play in a park. For the kids below the age of 5 years, they want to have a slide whose top is at a height of 1.5 m, and is disposed at an angle of 30° to the ground, while for older children, they prefer to have a steep slide based at a height of 3 m, and inclined at an angle of 60° to the ground. Find out the length of the slide in both the cases?

                                     (image to be added soon)

Solution:

Let the length of the slide for children below 5years of age x and the length of the slide for older children be y.

Given that: AF = 1.5 m and BC = 3m

∠FEA = 30°and ∠CDB = 60°

In right ▲FAE, sin 30°= AF/EF = 1.5/x

½ = 1.5/x

Thus x= 3m

In right ▲CBD, sin 60°= BC/CD = 3/y

√3/2 = 3/y

Thus, y = 3*2 / √3 = 2√3m

Therefore the length of the slide for children below 5years of age is 3m and the length of the slide for older children is 2√3m. 

Example: An air balloon is flying above the ground at a height of 60 m. The string joined to the balloon is temporarily tied to a point on the ground. The inclination of the string with the ground level is 60°. What will be the length of the string, supposing that there is no slack in the string?

                                  (image to be added soon)

Solution: Given that: AB = 60m and ∠ACB = 60°

Let AC be the length of the string

Then, in the right ▲ABC, sin 30°= AB/AC

= √3/2 = 60/AC

AC= 60*2/√3/ * √3/√3

= 120* √3 / 3

= 40√3

Therefore, the total length of the string is 40√3m

The value of root 5, when reduced to 5 decimal points, is √5 = 2.23606. It has a place with a large list of irrational algebraic numbers. It has been sorted in light because of the fact that the square root of 5 can’t be described as a fraction and has an eternal number of decimals. Additionally, the specific value can never be found perfectly. In Mathematics, the square root of 5 is presented or written as √5. It is a positive number and also the value of √5 when multiplied by itself, gives you the prime number 5. To distinguish itself from a negative number with the same properties, it is called as the principal root of 5.

 

How to Find the Square Root of 5?

This question might be bothering you for quite some time now. The simplest way to find the square root of any number would be by using the division method. How to find the value of root 5? Follow the steps given below:

 

Step 1: The first step is to group the digits in pairs of two. You start from the unit that is in the unit place and move towards the left-hand side for a number before the decimal point. For the number after the decimal point, you group the first two numbers and move towards the right-hand side.

5. 00 00 00 00 

 

Step 2: In this step, you will have to pick the largest square number that is either equal to or lesser than the first number pair. Now take this number as the divisor and also note down the quotient. 

 

Step 3: Now, you subtract the final product of the quotient and the divisor and the quotient from the pair of numbers or the number. Next, you bring down the next pair of numbers.

 

Step 4: You now need to calculate the divisor. To do that, you’ll have to multiply the previous quotient by 2 and then pick a new number in such a way that the digit and the new divisor is less than or equal to the new dividend

 

Step 5: Repeat Step 2, Step 3, and Step 4, until all the pairs of numbers are exhausted. Now, the quotient that you’ve found is the square root. In case of the value of under root 5, this is how it is done. 

 

uploaded soon)

 

 

Therefore, the square root of 5 = 2.236

 

What is the Square Root of 5?

The value of root 5, when reduced to 5 decimal points, is 2.23606 and this is just the simplified version of the value. In addition to that, the actual value of root 5 can be equal to at least ten billion digits. 

 

Sample Questions

1. Using the division method, find the square root of the value 784.

Solution:

 

 

2. Using the division method, find the square root of the value 5329

Solution:

 

 

3. Find the square root of 66049

Solution:

 

 

Did you know?

• 5 is not a perfect square as the square root of 5 is not a whole number.

•  The square root of 5 in exponential form can be written as (5)½ or (5)0.5.

• When solving a problem having a square root of 5 it is advisable to take the value till 3 decimal points.

 

Is the square root of the number 5 a rational number or an irrational number?

First let us understand what rational and irrational numbers are. A rational number is a number that can be written in the form of a ratio between any two integer numbers. For example, the square root of 9 is equal to 3, which can also be written as 3/1.

 

Whereas, an irrational number is a number that can not be written in the form of a ratio between any two integer numbers. So, the square root of 5 which is equivalent to 2.23 up to its two decimal values, which is an irrational number.

 

Solved examples

1.  Suppose the sides of a square photo frame is 2.33 m in length. Find out the area of the photo frame and write the answer to its nearest roundoff.

Ans. Length of the side of the photo frame= 2.33 m

Area of square = (side)2

Substituting the value of the length to the above equation we get,

Area of photo frame= (2.33)2= 5.4289m2

Rounding it off we get 5 m2

 

2. The area of a square-shaped wall i
s 25m2. What is the length of one side of the wall? What is the perimeter of the wall?

Ans. Area of square = (side)2

Substituting the value of the area of the wall we get,

25 = (side)2

√25 = side = 5 meters.

Perimeter of square = 4 x side

= 4 x 5 = 20 meters.

 

3. What is the value of 15√5?

Ans. 15√5 = 15 x 2.236 = 33.54101

 

4. Evaluate the following problems:

1. 5√16+2√25-2√5

2. √5 – √1

3. 20√25 – 10√9 – 5√5         

Ans.

1. 5√16+2√25-2√5

= (5 x 4) + (2 x 5) + (2 x 2.236)

= 20 + 10 + 4.472

= 34.472

 

2. √5 – √1

= 2.236 – 1

= 1.236

 

3.20√25 – 10√9 – 5√5

= (20 x 5) – (10 x 3) – (5 x 2.236)

= 100 – 30 – 11.18

= 58.82

Refer to the solved examples to understand how the concept of the square root of 5 has been used. Learn how it is defined and calculated to get a better idea of this topic. By doing this, you can easily find out the square roots of other whole numbers easily.