Mathematics – Page 18

Total Surface Area of Cylinder, Volume of a Cylinder Formulas, Examples

December 26, 2022

Total Surface Area of a Cylinder: Let us carry out a small activity. Take a ring. We know that ring is circular in shape. Now place another ring on the first ring and so on. You will see that when you place such rings, the structure that you get is a cylinder, which is three dimensional. Let us learn about the volume of a Cylinder and Total Surface area of a Cylinder in this chapter.

Cylinder

Any solid which has two circular ends along with a curved surface area is said to be a cylinder. It has a curved surface in the middle. The base and the top surface are identical. That means the bases are always parallel and congruent to each other. Cylinder has no vertices.

Total Surface Area of Cylinder

The total surface area of cylinder is the area of the circles plus the area of the round section in the midle. The Surface Area (SA) is found with the formula: SA = 2πr² + 2πrh

Find the Total Surface Area of each figure. Remember, d = 2r. Type your answers in the comment box

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

Calculate the total surface area of a cylinder for the given problems and type your answers in the comment box.

Volume of Cylinder

The Volume of Cylinder is the product of the area of the base (B) times the height. The formula for the Volume of the Cylinder is V = B x H. Volume is expressed in cubic units , or units³

Find the Volume of each figure. Remember, d = 2r. Type your answers in the comment box

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

Calculate the volume of cylinder for the given problems and type your answers in the comment box.

Maths

HCF AND LCM WORD PROBLEMS

December 23, 2022

HCF and LCM Word Problems : In this section, we will learn how to solve word problems involving highest common factor and lowest common multiple.

HCF and LCM Word Problems

Problem 1 :

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together? (excluding the one at start)

Solution :

For example, let the two bells toll after every 3 secs and 4 secs respectively.

Then the first bell tolls after every 3, 6, 9, 12 seconds…
Like this, the second bell tolls after every 4, 8, 12 seconds…

So, if the two bell toll together now, again they will toll together after 12 seconds. This 12 seconds is nothing but the L.C.M of 3 seconds and 4 seconds

The same thing happened in our problem. To find the time, when they will all toll together, we have to find the L.C.M of (2, 4, 8, 6, 10, 12).

L.C.M of (2, 4, 8, 6, 10, 12) is 120 seconds = 2 minutes.

So, after every two minutes, all the bell will toll together.

For example, in 10 minutes, they toll together :

10/2 = 5 times

That is, after 2,4,6,8,10 minutes. It does not include the one at the start

Similarly, in 30 minutes, they toll together :

30/2 = 15 times

(excluding one at the start).

Problem 2 :

The traffic lights at three different road crossings change after every 48 sec, 72 sec and 108 sec respectively. If they all change simultaneously at 8:20:00 hrs, when will they again change simultaneously ?

Solution :

For example, let the two signals change after every 3 secs and 4 secs respectively.

Then the first signal changes after 3, 6, 9, 12 seconds…

Like this, the second signal changes after 4, 8, 12 seconds…

So, if the two signals change simultaneously now, again they will change simultaneously after 12 seconds. This 12 seconds is nothing but the L.C.M of 3 seconds and 4 seconds

The same thing happened in our problem. To find the time, when they will all change simultaneously, we have to find the L.C.M of (48, 72, 108).

L.C.M of (48,72,108) is 432 seconds = 7 min 12 sec

So, after every 7 min 12 sec, all the signals will change simultaneously.

At 8:20:00 hrs, if all the three signals change simultaneously, again they will change simultaneously after 7 min 12 sec. That is at 8:27:12 hrs.

Hence, three signals will change simultaneously at 8:27:12 seconds.

Problem 3 :

A merchant has 120 ltrs of and 180 ltrs of two kinds of oil. He wants the sell oil by filling the two kinds of oil in tins of equal volumes. What is the greatest of such a tin.

Solution :

The given two quantities 120 and 180 can be divided by 10, 20,… exactly. That is, both the kinds of oils can be sold in tins of equal volume of 10, 20,… ltrs.

But, the target of the question is, the volume of oil filled in tins must be greatest.

So, we have to find the largest number which exactly divides 120 and 180.That is nothing but the H.C.F of (120, 180)

H.C.F of (120, 180) = 60

The 1st kind 120 ltrs is sold in 2 tins of of volume 60 ltrs in each tin.

The 2nd kind 180 ltrs is sold in 3 tins of volume 60 ltrs in each tin.

Hence, the greatest volume of each tin is 60 ltrs.

Problem 4 :

Find the least number of soldiers in a regiment such that they stand in rows of 15, 20, 25 and form a perfect square.

Solution :

To answer this question, we have to find the least number which is exactly divisible by the given numbers 15,20 and 25.That is nothing but the L.C.M of (15, 20, 25)

L.C.M of (15, 20, 25) = 300

So, we need 300 soldiers such that they stand in rows of 15, 20 , 25.

But, it has to form a perfect square (as per the question)

To form a perfect square, we have to multiply 300 by some number such that it has to be a perfect square.

To make 300 as perfect square, we have to multiply 300 by 3.
Then, it is 900 which is a perfect square.

Hence, the least number of soldiers required is 900.

Problem 5 :

Find the least number of square tiles by which the floor of a room of dimensions 16.58 m and 8.32 m can be covered completely.

Solution :

We require the least number of square tiles. So, each tile must be of maximum dimension.

To get the maximum dimension of the tile, we have to find the largest number which exactly divides 16.58 and 8.32. That is nothing but the H.C.F of (16.58, 8.32).

To convert meters into centimeters, we have to multiply by 100.

16.58 ⋅ 100 = 1658 cm

8.32 ⋅ 100 = 832 cm

H.C.F of (1658, 832) = 2

Hence the side of the square tile is 2 cm

Required no. of tiles :

= (Area of the floor) / (Area of a square tile)

= (1658 ⋅ 832) / (2 ⋅ 2)

= 344,864

Hence, the least number of square tiles required is 344,864.

Problem 6 :

A wine seller had three types of wine. 403 liters of 1st kind, 434 liters of 2nd kind and 465 liters of 3rd kind. Find the least possible number of casks of equal size in which different types of wine can be filled without mixing.

Solution :

For the least possible number of casks of equal size, the size of each cask must be of the greatest volume.

To get the greatest volume of each cask, we have to find the largest number which exactly divides 403, 434 and 465. That is nothing but the H.C.F of (403, 434, 465)

The H.C.F of (403, 434, 465) = 31 liters

Each cask must be of the volume 31 liters.

Req. No. of casks is

= (403/31) + (434/31) + (465/31)

= 13 + 14 + 15

= 42

Hence, the least possible number of casks of equal size required is 42.

Problem 7 :

The sum of two numbers is 588 and their HCF is 49. How many such pairs of numbers can be formed ?

Solution :

Because the H.C.F is 49, the two numbers can be assumed as 49x and 49y

Their sum is 588. So, we have

49x + 49y = 588

Divide each side 49.

x + y = 12

We have to find the values of “x” and “y” such that their sum is 12.

The possibles pairs of values of (x, y) are

(1, 11), (2, 10), (3, 9), (4, 8), (5, 7), (6, 6)

Here, we have to check an important thing. That is, in the above pairs of values of (x, y), which are all co-primes ?

[Co-primes = Two integers are said to be co-primes or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1]

Therefore in the above pairs, (1, 11) and (5, 7) are the co-primes.

Hence, the number of pairs is 2.

Problem 8 :

The product of two numbers is 2028 and their H.C.F. is 13. Find the number of such pairs

Solution :

Since the H.C.F is 13, the two numbers could be 13x and 13y

Their product is 2028.

So, we have

(13x) ⋅ (13y) = 2028

169xy = 2028

Divide each side by 169.

xy = 12

We have to find the values of “x” and “y” such that their product is 12.

The possibles pairs of values of (x, y) are

(1, 12), (2, 6), (3, 4)

Here, we have to check an important thing. That is, in the above pairs of values of (x, y), which are all co-primes?

[Co-primes = Two integers are said to be co-primes or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1]

Therefore in the above pairs, (1, 12) and (3, 4) are the co-primes.

Hence, the number of pairs is 2

Problem 9 :

Lenin is preparing dinner plates. He has 12 pieces of chicken and 16 rolls. If he wants to make all the plates identical without any food left over, what is the greatest number of plates Lenin can prepare ?

Solution :

To make all the plates identical and find the greatest number of plates, we have to find the greatest number which can divide 12 and 16 exactly.

That is nothing but H.C.F of 12 and 16.

H.C.F of (12, 16) = 4

That is, 12 pieces of chicken would be served in 4 plates at the rate of 3 pieces per plate.

And 16 rolls would be served in 4 plates at the rate of 4 rolls per plate.

In this way, each of the 4 plates would have 3 pieces of chicken and 4 rolls. And all the 4 plates would be identical.

Hence, the greatest number of plates Lenin can prepare is 4

Problem 10 :

The drama club meets in the school auditorium every 2 days, and the choir meets there every 5 days. If the groups are both meeting in the auditorium today, then how many days from now will they next have to share the auditorium ?

Solution :

If the drama club meets today, again they will meet after 2, 4, 6, 8, 10, 12…. days.

Like this, if the choir meets today, again they will meet after 5, 10, 15, 20 …. days.

From the explanation above, If both drama club and choir meet in the auditorium today, again, they will meet after 10 days.

And also, 10 is the L.C.M of (2, 5).

Hence, both the groups will share the auditorium after ten days.

Problem 11 :

John is printing orange and green forms. He notices that 3 orange forms fit on a page, and 5 green forms fit on a page. If John wants to print the exact same number of orange and green forms, what is the minimum number of each form that he could print ?

Solution :

The condition of the question is, the number of orange forms taken must be equal to the number of green forms taken.

Let us assume that he takes 10 orange and 10 green forms.

10 green forms can be fit exactly on 2 pages at 5 forms/page. But,10 orange forms can’t be fit exactly on any number of pages.

Because, 3 orange forms can be fit exactly on a page. In 10 orange forms, 9 forms can be fit exactly on 3 pages and 1 form will be remaining.

To get the number of forms in orange and green which can be fit exactly on some number of pages, we have to find L.C.M of (3,5). That is 15.

15 orange forms can be fit exactly on 5 pages at 3 forms/page.

15 green forms can be fit exactly on 3 pages at 5 forms/page.

Hence,the smallest number of each form could be printed is 15.

Problem 12 :

Lily has collected 8 U.S. stamps and 12 international stamps. She wants to display them in identical groups of U.S. and international stamps, with no stamps left over. What is the greatest number of groups Lily can display them in ?

Solution :

To make all the groups identical and find the greatest number of groups, we have to find the greatest number which can divide 8 and 12 exactly.

That is nothing but H.C.F of 8 and 12.

H.C.F of (8, 12) = 4

That is, 8 U.S stamps can be displayed in 4 groups at 2 stamps/group.

And 12 international stamps can be displayed in 4 groups at 3 stamps/group.

In this way, each of the 4 groups would have 2 U.S stamps and 3 international stamps. And all the 4 groups would be identical.

Hence, the greatest number of groups can be made is 4

Problem 13 :

Abraham has two pieces of wire, one 6 feet long and the other 12 feet long. If he wants to cut them up to produce many pieces of wire that are all of the same length, with no wire left over, what is the greatest length, in feet, that he can make them ?

Solution :

When the two wires are cut in to small pieces, each piece must of same length and also it has to be the possible greatest length.

6 feet wire can be cut in to pieces of (2, 2, 2) or (3, 3)

12 feet wire can be cut in to pieces of (2, 2, 2, 2, 2, 2 ) or (3, 3, 3, 3)

The length of each small piece must be of possible greatest length.

To find the possible greatest length, we have to find the greatest number which can divide both 6 and 12. That is H.C.F of (6, 12).

H.C.F of (6, 12) = 6.

Hence, the greatest length of each small piece will be 6 ft.

(That is, 6 feet wire is not cut in to small pieces and it is kept as it is. Only the 12 feet wire is cut in to 2 pieces at the length of 6 feet/piece)

Trig Cheat Sheet, Trig Identities, Inverse Trig Functions Cheat Sheet

December 15, 2022

Trig Cheat Sheet

Trig Cheat Sheet: Trigonometry is the study of triangles, which contain angles, of course. Get to know some special rules for angles and various other important functions, definitions, and translations. Sines and cosines are two trig functions that factor heavily into any study of trigonometry; they have their own formulas and rules that you’ll want to understand if you plan to study trig for very long.

Definition of the trig functions

Right Triangle Definition

Unit Circle Definition

Domain

Range | Trig Cheat Sheet

Period | Trig Cheat Sheet

Tangent and Cotangent Identities | Trig Identities Cheat Sheet

Reciprocal Identities | Trig Identities Cheat Sheet

Pythagorean Identities

Even/Odd Formulas

Periodic Formulas

Double Angle Formulas

Degree to Radians Formulas

As you study trig, you’ll find occasions when you need to change radians to degrees, or vice versa. A formula for changing from degrees to radians or radians to degrees is:

Half Angle Formulas

Sum and Difference Formulas

Product to Sum Formulas Cheat Sheet

Sum to Product Formulas Cheat Sheet

Cofunction Formulas Cheat Sheet

Unit Circle

Inverse Trig Functions

Trig Cheat Sheet PDF

Many of the formulas used in trigonometry are also found in algebra, calculus and analytic geometry. But trigonometry also has some special formulas usually found just in those discussions. A formula provides you a rule or equation that you can count on to work, every single time. Trigonometry formulas gives a relationship between particular quantities and units. The main trick to using formulas is to know what the different letters represent. In the formulas given here, you have: r (radius); d(diameter or distance); b (base or measure of a side); h (height); a, b, c (measures of sides); x, y (coordinates on a graph); m (slope); M (midpoint); h, k (horizontal and vertical distances from the center); θ (angle theta); and s (arc length). The formulas particular to trigonometry have: sin (sine), cos (cosine), and tan (tangent), although only sin is represented here.

Trigonometry Table

December 14, 2022

Trigonometry Table: Trigonometry is a branch of Mathematics deals with the study of length, angles and their relationships in a triangle. Trigonometric ratios are applicable only for right angle triangles, with one of the angle is equal to 90^o

Trigonometry Table

The trig table is made up of the following of trigonometric ratios that are interrelated to each other – sin, cos, tan, cos, sec, cot.

sin (reciprocal of cosecant) = opposite over hypotenuse
cos (reciprocal of secant) = adjacent over hypotenuse
tan (reciprocal of cotangent) = opposite over adjacent
cot (reciprocal of tangent) = adjacent over opposite
cosec (reciprocal of sine) = hypotenuse over opposite
sec (reciprocal of cosine) = hypotenuse over adjacent

The calculations can easily be figured out by memorizing a table of functions most commonly known as the Trigonometric Table. This find use in several areas. Some of them include navigation video games, aviation, science, geography, engineering, geometry etc. The trigonometric table helped in many developments and in the field of Mechanical Engineering for first innovation.

The Trig ratios table gives us the values of standard trigonometric angles such as 0°, 30°, 45°, 60°, and 90°. These values hold increased precedence as compared to others as the most important problems employ these ratios. It is therefore very important to know and remember the ratios of these standard angles.

Tricks to Remember Trigonometry Table

Remembering the trigonometry table will be useful as it finds many applications, and there are many methods to remember the table. Knowing the Trigonometry formulas, ratios and identities automatically will lead to figuring out the table and the values. The Trigonometric ratio table is depended upon the trigonometry formulas in the same way all the functions of trigonometry are interlinked with each other.

Before attempting to begin, it is better to try and remember these values, and know the following trigonometric ratios of complementary angles.

sin x = cos (90∘−x)
cos x = sin (90∘−x)
tan x = cot (90∘−x)
cot x = tan (90∘−x)
sec x = cot (90∘−x)
cot x = sec (90∘−x)

Reciprocal relations of Trigonometric Ratios

1 / sin x = cosec x
1 / cos x= sec x
1 / sec x= cos x
1 / tan x= cot x
1 / cot x= tan x
1/ cosec x = sin x

Steps to Create Trigonometric Table:

Step 1: Draw a tabular column with the required angles such as 0, 30^o, 45^o, 60^o, 90^o, 180^o, 270^o, 360^o in the top row and all 6 trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent in first column.

Step 2:

Step 3:

Step 4:

Step 5:

Step 6:

Step 7:

Applications of Trigonometry

December 14, 2022

Applications of Trigonometry: Trigonometry is the branch of mathematics deals with triangles involved with the study of lengths, heights, and angles. Trigonometry and its functions have an enormous number of uses in our real life.

Applications of Trigonometry

You have studied in social studies that the highest mountain peak in the world is Mount Everest and its height is 8848 meters.

Kuntala waterfall in Adilabad district is the highest natural waterfall in Andhra Pradesh. Its height is 147 feet.

How were these heights measured? Can you measure the height of your school building or the tallest tree in or around your school?

Let us understand through some examples. Vijaya wants to find the height of a palm tree. She tries to locate the
top most point of the tree. She also imagines a line joining the top most point and her eye.

This line is called “line of sight”. She also imagines a horizontal line, parallel to earth, from her eye to the tree.

Here, “the line of sight”, “horizontal line” and “the tree” form a right angle triangle.

To find the height of the tree, she needs to find a side and an angle in this triangle.

“The line of sight is above the horizontal line and angle between the line of sight and the horizontal line is called angle of elevation”.

Suppose you are standing on the top of your school building and you want to find the distance of borewell from the
building on which you are standing. For that, you have to observe the base of the borewell. Then, the line of sight from your eye to the base of borewell is below the horizontal line from your eye. Here, “the angle between the line of sight and horizontal line is called angle of depression.”

Trigonometry has been used by surveyors for centuries. They use Theodolites to measure angles of elevation or depression in the process of survey. In nineteenth century, two large Theodolites were built by British India for the surveying project“great trigonometric survey”. During the survey in 1852, the highest mountain peak in the world was discovered in the Himalayas. From the distance of 160 km, the peak was observed from six different stations and the height of the peak was calculated. In 1856, this peak was named after Sir George Everest, who had commissioned and first used the giant Theodolites. Those theodolites are kept in the museum of the Survey of India in Dehradun for display.

Drawing Figures to Solve Problems

When we want to solve the problems of heights and distances, we should consider the following:

All the objects such as towers, trees, buildings, ships, mountains etc. shall be considered as linear for mathematical convenience.
The angle of elevation or angle of depression is considered with reference to the horizontal line.
The height of the observer is neglected, if it is not given in the problem.

When we try to find heights and distances at an angle of elevation or depression, we need to visualise geometrically. To find heights and distances, we need to draw figures and with the help of these figures we can solve the problems. Let us see some examples.

Trigonometry in Real Life:

Electric poles are present everywhere. They are usually erected by using a metal wire. The pole, wire and the ground form a triangle. But, if the length of the wire decreases, what will be the shape of the triangle and what will be the angle of the wire with the ground ?
A person is whitewashing a wall with the help of a ladder which is kept as shown in the adjacent figure on
left. If the person wants to paint at a higher position, what will the person do? What will be the change in angle of the ladder with the ground ?
In the temple at Jainath in Adilabad district, which was built in 13th century, the first rays of the Sun fall at the feet of the Idol of Suryanarayana Swami in the month of December. There is a relation between distance of Idol from the door, height of the hole on the door from which Sun rays are entering and angle of sun rays in that
month. Is there any triangle forming in this context?
In a play ground, children like to slide on slider and slider is on a defined angle from earth. What will happen to the slider if we change the angle? Will children still be able to play on it?

The above examples are geometrically showing the application part of triangles in our daily life and we can measure the heights, distances and slopes by using the properties of triangles. These types of problems are part of ‘trigonometry’ which is a branch of mathematics.

Now look at the example of a person who is white washing the wall with the help of a ladder as shown in the previous figure. Let us observe the following conditions.

We denote the foot of the ladder by A and top of it by C and the point of joining height of the wall and base of the ladder as B. Therefore, ∆ABC is a right angle triangle with right angle at B. The angle between ladder and base is said to be θ.

If the person wants to white wash at a higher point on the wall-
● What happens to the angle made by the ladder with the ground?
● What will be the change in the distance AB?
If the person wants to white wash at a lower point on the wall-
● What happens to the angle made by the ladder with the ground?
● What will be the change in the distance AB?

We have observed in the above example of a person who was white washing. When he wants to paint at higher or lower points, he should change the position of ladder. So, when ‘θ’ is increased, the height also increases and the base decreases. But, when θ is decreased, the height also decreases and the base increases. Do you agree with this statement?

Here, we have seen a right angle triangle ABC and have given ordinary names to all sides and angles. Now let’s name the sides again because trigonometric ratios of angles are based on sides only

It may not have direct applications in solving practical issues but used in the various field. For example, trigonometry is used in developing computer music: as you are familiar that sound travels in the form of waves and this wave pattern through a sine or cosine function for developing computer music. Here are few applications where trigonometry and its functions are applicable. Trigonometric and its functions are applicable to measure the height of a building or a mountain, video games, constructions, aviation, physics, criminology, marine biology, marine engineering and navigation.

The calculus is based on trigonometry and algebra. The fundamental trigonometric functions like sine and cosine are used to describe the sound and light waves, Trigonometry uses in oceanography to calculate heights of waves and tides in oceans, creation of maps and satellite systems.

Thus here we have discussed Trigonometry and its importance as every student of math is expected to know about the application of this branch of mathematics in daily life. Solve sample questions with answers and cross check your answers with the NCERT Solutions on some applications of Trigonometry.

Trigonometric Ratios Table, Formulas, Definitions, Mnemonics, Problems

December 14, 2022

Six trigonometric ratios for right angle triangle are Sine(sin), Cosecant(Cos), Tangent(Tan), Cosecant(Cos), Secant(Sec), Cotangent(Cot) respectively. We will learn the sin, cos, tan formulas for these trigonometric ratios and easy ways to memorize it.

Trigonometry definition: Branch of Mathematics which deals with the measurement of Sides and angles of a triangle and the problems based on them.

Trigonometric Ratios

The ratios of the sides of the sides of a right triangle with respect to its acute angles.

Let us take a right triangle APM as shown in Figure. Here, ∠PAM (or, in brief, angle A) is an acute angle. Note the position of the side PM with respect to angle A. It faces ∠ A. We call it the side opposite to angle A. AP is the hypotenuse of the right triangle and the side AM is a part of ∠ A. So, we call it the side adjacent to angle A.

∠ A = θ, AP = r (Hypotenuse) and PM = y (Perpendicular), AM = x (Base), ∠ PMA = 90^o

Angle: A figure generated by rotating a given ray along of its end point.

Measurement of an Angle: Amount of rotation of the ray from initial position to the terminal position.

Hypotenuse Definition: the longest side of a right-angled triangle, opposite the right angle.

Perpendicular: at an angle of 90° to a given line, plane, or surface or to the ground.

Base: Side on which right angle triangle stands is known as its base

The Trigonometry Ratios of the angle θ in the triangle APM are defined as follows.

Opposite over Hypotenuse – Sin, Adjacent Over Hypotenuse – Cos, Opposite over Adjacent – Tan, Hypotenuse over Opposite – Cosec, Hypotenuse Over Adjacent – Sec and Adjacent over Opposite – Cotangent,

The ratios defined above are abbreviated as sin θ, cos θ, tan θ, cosec θ, sec θ and cot θ respectively. Note that the ratios cosec θ, sec θ and cot θ are respectively, the reciprocals of the ratios sin θ, cos θ and tan θ. So, the trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the length of its sides.

Opposite of Sin: Cosecant

Opposite of Cos: Secant

Opposite of Tan: Cotangent

Opposite of Cosecant: Sin

Opposite of Cotangent: Tan

Opposite of Secant: Cosecant

Trig Mnemonics – Some People Have, Curly Black Hair Through Proper Brushing.

Here, Some People Have is for

Sin θ= Perpendicular/ Hypotenuse.

Curly Black Hair is for

Cos θ= Base/ Hypotenuse.

Through Proper Brushing is for

Tan θ= Perpendicular/Base

Trigonometric Ratios of Some Specific Angles

We already know about isosceles right angle triangle and right angle triangle with angles 30º, 60º and 90º.
Can we find sin 30º or tan 60º or cos 45º etc. with the help of these triangles?
Does sin 0º or cos 0º exist?

Trigonometric Ratios of Angles Sin Cos Tan Chart — **Trig Table**

Trig Ratios of Complementary Angles

We know complementary angles are pair of angles whose sum is 90° Like 40°and 50°; 60°and 30°; 20°and 70°; 15° and 75° ; etc.

sin (90° – θ) = cos θ
cot (90° – θ) = tanθ
cos (90° – θ) = sin θ
sec (90° – θ) = cosec θ
tan (90° – θ) = cot θ
cosec (90° – θ) = sec θ

Origin of Trigonometric Ratios

The first use of the idea of ‘sine’ in the way we use it today was in the work Aryabhatiyam by Aryabhata, in A.D. 500. Aryabhata used the word ardha-jya for the half-chord, which was shortened to jya or jiva in due course. When the Aryabhatiyam was translated into Arabic, the word jiva was retained as it is. The word jiva was translated into sinus, which means curve, when the Arabic version was translated into Latin. Soon the word sinus, also used as sine,
became common in mathematical texts throughout Europe. An English Professor of astronomy Edmund Gunter (1581–1626), first used the abbreviated notation ‘sin’.

The origin of the terms ‘cosine’ and ‘tangent’ was much later. The cosine function arose from the need to compute the sine of the complementary angle. Aryabhatta called it kotijya. The name cosinus originated with Edmund Gunter. In 1674, the English Mathematician Sir Jonas Moore first used the abbreviated notation ‘cos’.

Sin Cos Tan are the main functions used in Trigonometry and are based on a Right-Angled Triangle.

Solved Examples on Trig Ratios:

Example-1. If tan A = 3/4 , then find the other trigonometric ratio of angle A.
Solution :

Given tan A = 3/4
Hence tan A = Opposite side/Adjacent side = 3/4
Therefore, opposite side : adjacent side = 3:4
For angle A, opposite side = BC = 3k
Adjacent side = AB = 4k (where k is any positive number)
Now, we have in triangle ABC (by Pythagoras theorem)

Example 2: If ∠ A and ∠ P are acute angles such that sin A = sin P then prove that ∠ A = ∠ P
Solution : Given sin A = sin P

Example 3: In ∆ABC, right angle is at B, AB = 5 cm and ∠ACB = 30^o. Determine the lengths of the sides BC and AC.
Solution: Given AB=5 cm and
∠ACB=30^o.

To find the length of side BC, we will choose the trignometric ratio involving BC and the given side AB. Since BC is the side adjacent to angle C and AB is the side opposite to angle C.
Therefore,
AB/BC = tan C

Example 4: A chord of a circle of radius 6cm is making an angle 60^o at the centre. Find the length of the chord.
Solution: Given the radius of the circle OA = OB = 6cm
∠ AOB = 60^o

Example-5. In ∆PQR, right angle is at Q, PQ = 3 cm and PR = 6 cm. Determine ∠QPR and ∠PRQ.
Solution : Given PQ = 3 cm and PR = 6 cm

Note : If one of the sides and any other part (either an acute angle or any side) of a right angle triangle is known, the remaining sides and angles of the triangle can be determined.

You can easily remember all trigonometry formulas using super magical hexagon, great way to remember all formulas easily.

Maths

December 14, 2022

Maths is a subject which is not only applicable in our academics but also in day to day life. Making Students love this subject games is a difficult challenge, especially for parents. Learning Mathematics takes a lot of brainpower to master, and this can be tough for students. This subject has a lot of formulas based on different topics. Maths formulas can be memorized by practicing questions based on them. Some problems can be solved quickly using Maths Shortcuts. Class 1 to 10 has been taught with the general math concepts, but it’s level increases in Intermediate class 11 and 12 (plus one and plus two maths).

At the starting level, basics of Math has been taught such as Introduction to Numbers and counting, addition, subtraction, multiplication, division, place value, face value etc. As the level of grade increases, students are taught with more enhanced concepts, such as Percentage,Ratio and Proportion, fractions, algebra, geometry, trigonometry, mensuration, etc. Integration and differentiation are the higher level of topics in class 11 and class 12, which are included in the syllabus of higher secondary school. Get Maths syllabus for class 9 to 12, here and prepare your studies.

Learn Basic Maths, Topics and Formulas for Mathematics

Algebra

Introduction to Algebra

Algebra Basics
Addition And Subtraction Of Algebraic Expressions
Multiplication Of Algebraic Expressions
BODMAS And Simplification Of Brackets
Substitution Method
Solving Inequalities

Exponents

Introduction to Exponents
Square Roots and Cube Roots
Surds
Simplifying Square Roots
Laws of Exponents
Exponents in Algebra

Simplifying

Associative Property, Commutative Property, Distributive Laws
Cross Multiply
Fractions in Algebra

Polynomials

What is a Polynomial?
Adding And Subtracting Polynomials
Multiplying Polynomials
Rational Expressions
Dividing Polynomials
Polynomial Long Division
Conjugate
Rationalizing The Denominator

Quadratic Equations

Solving Quadratic Equations
Factoring Quadratics
Completing the Square.

Arithmetic | Maths

Arithmetic Progression
Addition and subtraction
Laws of exponents
Multiplication and division
Decimals
Comparing Fractions
Percentages
Ratios and Proportion
Like and Unlike Fractions
Surds
Sum of n terms
Integers
Geometric Sequence
Fundamental Theorem
Fraction to decimal
Variable
Number Line
Whole Number
Even and odd numbers
Word problems on Decimals
Integers on a number line

Trigonometry

Calculus | Maths

Limits and Continuity
Limits
Continuity
Continuity and Differentiability
Mean Value Theorem
Second Derivative Test
Applications of Integration
Discontinuity
Partial Differential Equations
Application of Derivatives
Trapezoidal Rule
Differential Calculus
Derivatives
Derivative of a function
Quotient Rule
Chain Rule
Anti-derivative Formula
Integration by Parts
Cauchy Riemann Equations
Exact Differential Equation
L Hospital Rule
Line Integral
Differentiation and integration
Integral Calculus
Integration
Methods of Integration
Definite Integral
Indefinite Integral
Derivative of Inverse Trigonometric Function
Integrating Factor
Riemann sums
Simpson’s Rule
Surface Integral

Learning Maths has never been easier. At LearnCBSE, you can assure to create such a strong foundation in this subject, wall maths problems would seem to be a cakewalk to you. Getting a good grip in Maths fundamentals, through understanding and practice, will tend to develop your brain in a logical way and help you to grasp the topics in all other subjects with greater speed, accuracy and efficiency. In Maths Practice is key so spend most of the time in practicing, LearnCBSE Provides you Maths Formulas and Exercise Problems.

LearnCBSE providing learning materials for students of class 1 to 12 based on the syllabus, where students set their foot into the world of all the important ideas in Maths, like Natural Numbers, whole numbers, basic geometry, integers, decimals, algebra, ratio and proportion, trigonometry etc. Since kids are more receptive to learning when it is associated more with play than with work, it is a good idea for parents to expose kids to math in fun ways. We provide lessons for students till class 12, where they study the most complex topics like 3-D geometry, vector algebra, differential equations, Limits, Sets, Determinant, etc. Some of the links of important topics have been shared here for your convenience.

With the help of animations and innovative ways of teaching by our experts, the ideas and concepts are sure to stay etched in your minds forever. Making math fun and help kids grow to understand that math is fun, Instead of being afraid, the kids will be excited about the subject for the rest of their educational journey. You can engage yourself with interactive video sessions, regular tests prepared by our experts and do continuous analysis based on your performance.

How to Find Areas and Perimeters of Plane Figures

November 27, 2022

How to Find Areas and Perimeters of Plane Figures

Recommended Reading On: Java Program to Find Perimeter of Triangle

How to Represent the Data in Statistics

November 18, 2022

Representation of a Set | Statement Form, Roster Form, Set Builder Form

July 7, 2022

Set is a well-defined collection of objects or elements. A Set is represented using the Capital Letters and the elements are enclosed within curly braces {}. Refer to the entire article to know about Representation of Set in three different ways such as Statement Form, Set Builder Form, Roster Form. For a Complete idea on this refer to the Set Theory and clear all your queries. Check out Solved Examples for all three forms explained step by step.

Representation of a Set

Sets can be represented in three different forms. Let us discuss each of them in detail by taking enough examples.

Statement Form or Descriptive Form
Roster or Tabular Form
Rule or Set Builder Form

Statement Form: In this Form, well-defined descriptions are provided for the elements or members in the set. Verbal Description of Elements is given. Statement Form is also known as Descriptive Form. Elements of the set are enclosed within curly brackets {}.

Example: Set of Odd Numbers Less than 10.

In Statement Form, the elements can be expressed as {1, 3, 5, 7, 9}

Set of Students in Class V having a height above 5 ft.

Set of Numbers greater than 30 and less than 40.

Roster Form: In the Roster Form elements of the set are represented within {} and are separated by commas. In this representation order of the elements doesn’t matter but the elements must not be repeated. Roster Form is also known as Tabular Form.

Example:

1. Set of Natural Numbers less than 10

Set of Natural Numbers Less than 10 = {1, 2, 3, 4, 5, 6, 7, 8, 9}

Set N in Roster Form is {1, 2, 3, 4, 5, 6, 7, 8, 9}

2. Set of Natural Numbers that divide 10

Y = { 1, 2, 5, 10}

3. W is the Set of Vowels in the Word Elephant

W = {E, A}

Kbps Full Form – Kilobits per second · kbps stands for kilobits per second.

Set Builder Form: In this form, all the elements possess a single property in order to be members of the set. In this representation of the set, the element is denoted by the symbol x or any variable followed by the symbol : or |. After this symbol write down the property possessed by the elements of the set and enclose the entire description within brackets.

Example:

1. Write the following Set in Set Builder Form = { 3, 6, 9, 12}

Set Builder Form is A = {x: x= multiples of 3, n ∈ N and n ≤ 15}

2. D = {x: x is an integer and – 2 < x < 11}

3. X = {m: m is a negative integer < -10}

Percentage Decrease Calculator | Calculating Percentage Change

July 6, 2022

Wondering how to calculate Percentage Decrease then this is the article for you. We have curated the Formula to Calculate Percentage Decrease, Procedure to find % Decrease manually by hand in detail. Check out the Solved Examples for a better understanding of the concept. You can solve various questions regarding the concept and apply the concept in your day to day life.

Percentage Decrease Formula

The Formula to Calculate Percentage Decrease is given as under

Percentage Decrease = (Decreased Value/Original Value)*100

Decreased Value = Original Value – New Value

% Decrease = ((Original Value – New Value)/Original Value)*100

Have a look at the Percentage Decrease Example Questions provided below to get a grip on the concept.

How to Calculate the Percentage Decrease?

Go through the below-listed guidelines to find the Percentage Decrease. They are as such

Firstly find the Decreased Value i.e. Original Value – New Value.
Divide it with the Original Value.
Multiply the fraction with 100 and place a % sign at the end.

RSD Calculator is a small, quick & reliable and it works on all Android devices with Android 1.6 and up.

Example Questions on Percentage Decrease

1. Price of Sugar Decreases from $9 to $7.5 per Kg? What is the Percentage Decrease in the Price of Sugar?

Solution:

Original Price of the Sugar = $9

New Price of the Sugar = $7.5

Decreased Price of Sugar = $9- $7.5

= $1.5

Percentage Decrease = (Decreased Value/Original Value)*100

= ($1.5/$9)*100

= 16.66 %

Therefore, the Percentage Decrease in the Price of Sugar is 16.66%.

2. A shopkeeper sells a pair of pens for Rs. 30 initially. He then reduced the price of the pair of pens to Rs. 24. Calculate the Percentage Decrease in the cost of pens?

Solution:

Original Cost of Pair of Pens = Rs. 30

New Cost of Pair of Pens = Rs. 24

Decreased Value = Original Cost – New Cost

= Rs. 30 – Rs. 24

= Rs. 6

Percentage Decrease = (Decreased Value/Original Value)*100

= (6/30)*100

= 20%

Therefore, Percentage Decrease in the cost of pens is 20%.

3. Find the Decreased Value if 400 decreased by 25%?

Solution:

New Value 25%(400)

= 25/100*400

= 100

Decreased Value = Original Value – New Value

= 400 – 100

= 300

4. A fruit seller used to sell bananas for Rs. 30 per dozen. Later, he reduced the cost of a dozen bananas by 15%. Find the price of a dozen bananas now?

Solution:

New Prize = 15% of 30

= 15/100*30

= 450/100

= 4.5

Decreased Value = Original Price – New Prize

= 30- 4.5

= 25.5

Fruit Seller sells bananas Rs. 25.5/- a Dozen.

5. The membership card of a club cost was reduced by 10% and costs Rs. 450 now. What was the original price of the membership card before its cost is reduced?

Solution:

New Value = 10% of 450

= 10/100*450

= 45

Membership Card of a Club Cost initially = Reduced Price + Cost of Membership Card of Club after Reducing

= 45+ 450

= 495

The original Price of a membership card of the club is Rs. 495.

Category: Mathematics

Cylinder

Total Surface Area of Cylinder

Volume of Cylinder

HCF and LCM Word Problems

Trig Cheat Sheet

Right Triangle Definition

Unit Circle Definition

Domain

Range | Trig Cheat Sheet

Period | Trig Cheat Sheet

Tangent and Cotangent Identities | Trig Identities Cheat Sheet

Reciprocal Identities | Trig Identities Cheat Sheet

Pythagorean Identities

Even/Odd Formulas

Periodic Formulas

Double Angle Formulas

Degree to Radians Formulas

Half Angle Formulas

Sum and Difference Formulas

Product to Sum Formulas Cheat Sheet

Sum to Product Formulas Cheat Sheet

Cofunction Formulas Cheat Sheet

Unit Circle

Inverse Trig Functions

Trigonometry Table

Tricks to Remember Trigonometry Table

Reciprocal relations of Trigonometric Ratios

Steps to Create Trigonometric Table:

Applications of Trigonometry

Trigonometry in Real Life:

Trigonometric Ratios

Trigonometric Ratios of Some Specific Angles

Trig Ratios of Complementary Angles

Origin of Trigonometric Ratios

Solved Examples on Trig Ratios:

Learn Basic Maths, Topics and Formulas for Mathematics

Algebra

Introduction to Algebra

Exponents

Simplifying

Polynomials

Quadratic Equations

Arithmetic | Maths

Trigonometry

Calculus | Maths

How to Find Areas and Perimeters of Plane Figures

Representation of a Set

Percentage Decrease Formula

How to Calculate the Percentage Decrease?

Example Questions on Percentage Decrease