Category: Maths Notes PPT

An equation of the form ax2 + bx + c = 0, where a, b, c are real numbers and a0 is called a quadratic equation. The value of unknown variable x, which satisfies the given quadratic equation is called the roots of quadratic equation. For example, if is a root of quadratic equation ax2 + bx + c = 0, then a2 + b + c = 0. And the process of finding roots is known as solving a quadratic equation. There are three methods to solve a quadratic equation, which are as follows:

1. Solving a quadratic equation by factorisation.

2. Solving a quadratic equation by completing the square.

3. Solving a quadratic equation using quadratic formula.

In this article, we will learn about quadratic formula, its derivation and how to solve a quadratic equation using quadratic formula. 

What is Quadratic Formula?

Quadratic formula is a formula that helps us to find the roots of a quadratic equation very easily by replacing the other methods of finding the roots like, factorisation method, completing the square method. The quadratic formula to find the roots of quadratic equation ax2 + bx + c = 0, where a ≠ 0 is given by:

[x = frac{-b pm sqrt{b^{2} – 4ac}}{2a}]

The plus (+) and minus (-) sign represents that quadratic formula will give two roots, one root corresponding to the plus (+) sign and another root corresponding to the minus (-) sign.

i.e.,  [x_{1} = frac{-b – sqrt{b^{2} – 4ac}}{2a}] and [x_{2} = frac{-b + sqrt{b^{2} – 4ac}}{2a}]. Both roots are evaluated by substituting the corresponding values of coefficients a, b and c from the quadratic equation ax2 + bx + c = 0.

Derivation of Quadratic Formula

Consider the quadratic equation ax2 + bx + c = 0, where a, b, c are real numbers and a0. Then,

ax2 + bx + c = 0

⇒ ax2 + bx = –c

On dividing both sides by a, we get:

⇒ [x^{2} + frac{b}{a}x = frac{-c}{a}]

Now adding [(frac{b}{2a})^{2}] on both sides, we get:

⇒ [x^{2} + frac{b}{a}x + (frac{b}{2a})^{2} = frac{-c}{a} + (frac{b}{2a})^{2}]

⇒ [(x  + frac{b}{2a})^{2} = (frac{-c}{a} + frac{b^{2}}{4a^{2}})]

⇒ [(x + frac{b}{2a})^{2} = frac{(b^{2} – 4ac)}{4a^{2}}]

⇒ [(x – frac{b}{2a}) = frac{pm sqrt{b^{2} – 4ac}}{2a}], when (b2 – 4ac) 0

⇒ [x = – frac{b}{2a} pm frac{sqrt{b^{2} – 4ac}}{2a}]

⇒ [x = frac{-b pm sqrt{b^{2} – 4ac}}{2a}]

Therefore, the roots of quadratic equation ax2 + bx + c = 0 is  [x_{1} = frac{-b + sqrt{b^{2} – 4ac}}{2a}] and [x_{2} = frac{-b – sqrt{b^{2} – 4ac}}{2a}].

Discriminant of Quadratic Equation

The expression D = (b2 – 4ac) is called the discriminant of quadratic equation ax2 + bx + c = 0. 

The roots of ax2 + bx + c = 0 are real only when D 0 i.e., b2 – 4ac 0 and the roots are given by [x_{1} = frac{-b + sqrt{D}}{2a}] and [x_{2} = frac{-b – sqrt{D}}{2a}].

How to Solve Quadratic Equations using Quadratic Formula?

To learn how to solve quadratic equations using quadratic formula, let us consider some examples and solve them using quadratic formula.

Example 1: Find the roots of quadratic equation 15x2 – x – 28 = 0 using quadratic formula.

Solution:

The given quadratic equation is 15x2 – x – 28 = 0. Comparing it with ax2 + bx + c = 0, we get a = 15, b = -1 and c = -28.

So, D = b2 – 4ac = (-1)2 – 4 × 15 × (-28) = 1681. As D = 1681 > 0, The given quadratic equation has real roots.

Now substituting the corresponding values of a, b and c in quadratic formula: [x = frac{-b pm sqrt{b^{2} – 4ac}}{2a}], we get,

[x = frac{-(-1) pm sqrt{(-1)^{2} – 4 times 15 times (-28)}}{2 times 15}]

⇒ [x = frac{1 pm sqrt{1681}}{30}]

For plus (+) sign, 

The root is [x_{1} = frac{1 + sqrt{1681}}{30} = frac{1 + 41}{30} = frac{42}{30} = frac{7}{5}].

and, for minus (-) sign, 

the root is [x_{2} = frac{1 – sqrt{1681}}{30} = frac{1 – 41}{30} = frac{-40}{30} = frac{-4}{3}]

Hence, the required roots of quadratic equation 15x2 – x – 28 = 0 are 7/5 and -4/3 .

Example 2: Find the roots of quadratic equation x2 + 6x + 6 = 0 using quadratic formula.

Solution:

The given quadratic equation is x2 + 6x + 6 = 0. Comparing it with ax2 + bx + c = 0, we get a = 1, b = 6 and c = 6.

So, D = b2 – 4ac = (6)2 – 4 × 1 × 6 = 12. As D = 12 > 0, The given quadratic equation has real roots.

Now substituting the corresponding values of a, b and c in quadratic formula: [x = frac{-b pm sqrt{b^{2} – 4ac}}{2a}], we get,

[x = frac{-(-6) pm sqrt{(6)^{2} – 4 times 1 times 6}}{2 times 1}]

⇒ [x = frac{-6 pm sqrt{12}}{2}]

For plus (+) sign, 

the root is [x_{1} = frac{-6 + sqrt{12}}{2} = frac{-6 + 2sqrt{3}}{2} = (-3 + sqrt{3})]

and, for minus (-) sign, 

the root is the root is [x_{2} = frac{-6 – sqrt{12}}{2} = frac{-6 – 2sqrt{3}}{2} = (-3 – sqrt{3})] 

Hence, the required roots of quadratic equation x2 + 6x + 6 = 0 are [((-3 + sqrt{3})] and  [(-3 – sqrt{3})].

The distance between two points is measured as the length of the interval which joins two points. If the two points lie on the same horizontal or same vertical line then the distance is being calculated by subtracting the coordinates which are different. In the analytical geometry, the distance formula is useful to compute the distance that measures between two lines also. It is also useful to find the sum of the lengths of all the sides of a polygon, an area of polygons, and many more. It is also useful to determine a given triangle as scalene, isosceles, or equilateral.

How to Find the Distance Between Two Points

The distance between the two points along the XY-plane can be found using the distance formula. An ordered pair (x, y) represents the coordinate of the point, where x-coordinate (or abscissa) is the distance of the point from the centre and y-coordinate (or ordinate) is the distance of the point from the centre.

The Distance Formula is a useful method to find the length between two points in any shape. We may represent these points using coordinate geometry. This formula itself is actually derived with the help of the Pythagorean Theorem.

This theorem establishes a relationship as bellow,

c2 = a2 + b2

Where c is the longest side of a right triangle means hypotenuse. And a and b are the other shorter sides means perpendicular and base. The main purpose of the Distance Formula is to compute the length of the hypotenuse of the right triangle represented as c.

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Explanation of the Distance Formula Between Two Points in a 2d Plane:

This derivation is based on the Pythagoras theorem. Let us consider two points P (x1, y1) and Q (x2, y2) on the given coordinate axis. Then the distance between these points will be as given:

D = [sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}]

The expression x2 – x1 is referred to as the difference of the x-coordinate values of both points. Also, the expression y2 – y1 referred to as the difference of the y-coordinate values of both points.

Explanation of The Distance Formula Between Two Points in a 3d Plane

AS in the 2d plane, now let us consider two points P (x1, y1, z1)and Q (x2, y2, z2)on the given coordinate axis of the 3d plane. Then the distance between these points will be as given:

D = [sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{^{2}}}]

Distance Formula of a Point Form a Given Line: 

It is the perpendicular distance of a point from a line. We can measure the distance of the point from a given line.

Let us consider the equation of a straight line is Ax + By + C = 0, where A, B, and C are the coefficients. Also, assume that the point P in the Cartesian coordinate system with coordinates (x1, y1). Then the distance between line and point will be the length of the perpendicular drawn from point P to the line.  

The distance formula from a point to a line is as given below:

d = [frac{|Ax_{1} + By_{1} + C|}{sqrt{A^{2} + B^{2}}}]

Minimum Distance Between Two Parallel Lines

The distance between parallel lines is the minimum distance from any point on one of the lines to the other line. This distance is defined as the distance between two parallel lines, and obviously, it will be the perpendicular distance between them.

Let us assume the two straight lines parallel to each other as:

The first line is,

Line-1 : y = mx + c1 , this equation is in slope-intercept form.

And the second line is

Line-2 : y = mx + c2 , this equation is in slope-intercept form.

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Let us consider that ‘d’ is the distance between both the lines.

Formula to find the minimum distance between two non-intersecting lines as given below:

d = [frac{|C_{1} – C_{2}|}{sqrt{A^{2} + B^{2}}}]

Distance Between Two Points in Polar Coordinates: 

We have used the Cartesian system for finding out the distance. Now, we will use the polar coordinate system for the same purpose. We know that in the polar coordinate system, the same point will be having polar coordinates in terms of r and

.

So, let us assume that O be the pole and OX be the initial line. Let P and Q be two given points whose polar coordinates are ( [r_{1}, theta_{1}] ), and ( [r_{2}, theta_{2}] ) respectively.  Now, the distance formula between the points is given by,

d = [sqrt{r_{1}^{2}+r_{2}^{2}-2r_{1}r_{2}cos(theta_{1}- theta_{2})}]

Also, here [theta_{1}, theta_{2}] are the angles measured in radians.

Fun Facts:

• It is a scalar quantity, as it does not depend on the direction. 

• Distance between two points depends on the coordinates of the endpoints.

• Distance between two points cannot be negative. Because negative value has no meaning.

• Distance between two points helps to prove many geometrical problems and principles as well.

• The distance formula is also applicable in polar coordinate systems.

The chord of a circle can be stated as a line segment joining two points on the circumference of the circle. The diameter is the longest chord of the circle which passes through the center of the circle. The figure shown below represents the circle and its chord.

In the circle above with center O, AB represents the diameter of the circle (longest chord of a circle), OE represents the radius of a circle and CD represents the chord of a circle.

Let us consider CD as the chord of a circle and points P and Q lying anywhere on the circumference of the circle. If the two endpoints of the chord CD meet at point P, then ∠CPD is known as the angle extends by the chord CD at point P. The angle ∠CQD is known as the angle extended by the chord CD at point Q. The angle ∠CPD is known as the angle extended by the chord CD at point P.

In this article, we will study what is a chord in a circle, chord length formulas, how to find the length of the chord, length of the common chord of two circles formulas, chord radius formulas, etc.

How to find the Length of the Chord?

There are two important formulas to find the length of the chords. The formula for the length of a chord is given as:

Chord Length Formula Using Perpendicular Distance from the Center 

Chord Length = [2 times sqrt{r^2 – d^2} ]

Chord Length Formula Using Trigonometry

Chord Length =[ 2 times  r times sin(frac{c}{2}) ]

In the above formula for the length of a chord,

R represents the radius of the circle

C represents the angle extended at the center by the chord.

D represents the perpendicular distance from the cord to the center of the circle.

Chord Radius Formula

The chord radius formula when length and height of the chord are given is

[R = frac{L^2}{8h} + frac{h}{2} ]

In the above chord radius formula,

R is the radius of a circle

L is the length of the chord

h is the height of th chord

Length of Common Chord of Two Circles Formula

The length of the common chord of two circles formulas when the radius of two circles and distance between the center of the two circles is given below.

Length of the common chord of the two circle formula is:

2 × radius 1 × radius 2 ÷ Distance between the center of two circles

Other Parts of a Circle

• The radius of Circle: The radius of a circle is described as the distance from the center to any other point on the boundary of the given circle. 

• Diameter of a Circle: The diameter of a circle is a straight line passing through the center of the circle that connects the 2 points on the boundary. The diameter of a given circle is always twice the radius of the given circle. 

• Circumference of Circle: The circumference of a circle is described as the perimeter of a circle. It is the distance around the boundary of the given circle. It is found by the formula- 

C = 2 × π × r (where r is the radius of the given circle)

• Area of a Circle – The area enclosed by a circle or the region that it occupies in a 2-Dimensional plane is called the area of the circle. Area of a given Circle) A = π × r2

• Arc of a Circle: An arc is the curve part or portion of a circle’s circumference. Mathematically it can be calculated as – s = 2 π r (θ /3600) (where s is the length of the arc, r is radius of the circle and θ is the central angle of the circle)

• The Secant of Circles:  The secant of a circle is the line that crosses 2 points on the circumference of a circle. The word ‘secant’ literally means ‘cut’ in Latin. There exists a secant-tangent rule that states that,  when a secant line and a tangent of a given circle are constructed from a common exterior point, the multiplication of the secant and its external segment is always equal to the square of the tangent.

• Tangent: Tangent is the line that cuts the border of the circle exactly at 1 single point. The tangent point doesn’t enter the interior of the circle. A circle can have infinite tangents as it is made up of infinite points. Tangents will always be perpendicular to the radius of the given circle.

• A Segment in a Circle: A segment of a circle is the area enclosed by a chord and the corresponding arc in a given circle. Segments are divided into 2 types -minor segment, and major segment. The Area of a segment = r2 (θ – sinθ) ÷ 2  (where r is the radius of the circle and θ is in radians)

•  A Sector of a Circle: The sector of a circle is the area enclosed by 2 radii and the corresponding arc. Sectors are divided into 2 types of sectors, minor sector, and major sector. If the sector cuts an angle θ (in radians) at the center, the area of a sector of that given circle = (θ x r2 ) ÷ 2 

Solved Examples

1. Calculate the length of the chord where the radius of the circle is 7cm and the perpendicular distance drawn from the center of the circle to its chord is 4 cm.

Solution:

Given,

Radius of a circle = 7 cm

And Distance, d = 4 cm

Length of Chord Formula Circle = [2 sqrt{r^2 – d^2}]

Chord Length = [2 sqrt{7^2 – 4^2}]

Chord Length = [2 sqrt{49 – 16}]

Chord Length = [2 sqrt{33}]

Chord Length = 2 [times] 5.744

Or chord length = 11.488 cm

2. In the circle given below, find the measure of ∠POQ when the value of ∠PRQ is given as 62°.

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Solution:

According to the property of chords of a circle, the angle extended at the center of the circle and an arc is twice the angle extended by it at any point on the circumference.

 

Hence, ∠POQ is equal to twice of ∠PRQ. Hence, [angle POQ = 2 times  sqrt{PRQ}]

∠POQ = 2 x 62° = 124°.

Quiz Time 

1. Find the length of the chord in the above- given circle

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1. 5

2. 4

3. 7

4. 6

2. The longest chord of the circle is

1. Radius

2. Diameter

3. Segment

4. Arc

3. If chords PQ and RS of congruent circles subtend equal angles at their centers, then:

1. PQ = RS

2. PQ > RS

3. PQ < RS

4. None of the above

While calculating problems based on coordinate geometry, you will come across the problems such as calculating the distance between two points, to find the equation, midpoint, and slope of the line segment, or even more complex problems that require the use of the coordinate geometry formulas. But have you ever thought, how to construct a polygon on a cartesian plane?

A polygon is an area enclosed by multiple straight lines with a minimum of three straight lines known as a triangle, to an infinite number of straight lines. Calculating the area of the triangle using the coordinates of its vertices is a much-discussed topic in geometry.

Here, we will discuss how to use coordinate formulas to calculate the area of the triangle using coordinates of its vertices.

What is Coordinate Geometry?

Coordinate Geometry, also called analytical geometry is a Mathematics subject in which algebraic methods and symbols are used to solve problems in Geometry. The importance of coordinate geometry is that it establishes the correlation between algebraic equations and geometric curves. This correlation helps to redevelop problems in Geometry as similar problems in Algebra, and vice versa. These methods can also be used to solve problems in other areas. For example, computers develop animations for display in games and films by using algebraic equations.

Area of Triangle in Coordinate Geometry

As we know, coordinate geometry is the study of geometry using the coordinate points. We can also determine the area of the triangle in coordinate geometry if the coordinates of the vertices of a triangle are given.

Example:

Consider the following points:

• A = (x₁, y₁)

• B  = ( x₂, y₂)

• C = (x₃, y₃)

If you plot these points in a plane, you will find that they are non – collinear, which means that they can be considered as the vertices of a triangle as shown below:

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In Coordinate geometry, we can determine the area of a triangle using coordinates of its vertices.

Method to Find Area of Triangle Using Coordinates

We can follow the following points if we want to find the area of the triangle using coordinates of its vertices.

1. Plot the given coordinates in a plane.

2. Locate the coordinates of the vertices of the triangle in a counter-clockwise direction, else the formula will give negative values.

3. Use the area of triangle formula given below.

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4. Add the product of the diagonals  (x₁, y₁), ( x₂, y₂), (x₃, y₃) as shown in the dark blue arrow.

5. Also, add the product of the diagonals (x₂, y₁), ( x₃, y₂), (x₁, y₃) as shown in the dark blue arrow.

6. Now, subtract the latter products of the diagonals with the former product to get the area of the triangle.

7. Hence, the area of a triangle using coordinates of its vertices can be calculated as:

½ {(x₁y₂ + x₂y₃ + x₃y₁) –  (x₂y₁ + x₃y₂ + x₁y₃)}

Area of Triangle Formula in Coordinate Geometry

If the coordinates of vertices of a triangle (x₁, y₁), ( x₂, y₂), (x₃, y₃) are given, then the area of triangle formula in coordinate geometry is given as:

½ {(x₁y₂ + x₂y₃ + x₃y₁) –  (x₂y₁ + x₃y₂ + x₁y₃)}

Let us understand the concept with an example.

Examples

1. Find the area of the triangle whose vertices are (-2,1), (3,2), and (1,5).

Solution:

First, we will plot the given points on the graph as shown below:

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As we have to take the points in anticlockwise directions, we will take the points in the order C ( -3,5) , A (-2,1),  and B (3,2).

x₁ = -3 x₂ = -2 x₃ = 3

y₁ = 5         y₂ = 1 y₃ = 2

The following pictorial representation helps us to calculate the area of the triangle easily.

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Area of △CAB = ½ {(x₁y₂ + x₂y₃ + x₃y₁) –  (x₂y₁ + x₃y₂ + x₁y₃)}

Substituting the values, we get

= ½ {(-3 × 1) + (-2 × 2) + (3 × 5) – (-2 × 5 ) + (3 × 1) + (-3 × 2)}

= ½ {( -3) + (-4) + (15)} –  (-10) + (3) + (-6)}

= ½ ( -3  – 4 + 15) –  (-10 + 3 + -6)

= ½ (8 + 13)

= ½ (21)

= 10.5 square units

Hence, the area of △CAB = 10.5 square units.

2. Find the area of the triangle whose vertices are (3,1), (0,4), and (-3,1).

Solution:

First, we will plot the given points on a graph as shown below:

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As we have to take the points in anticlockwise directions, we will take the points in the order B (0,4), A (-3,1),  and B (3,1).

x₁ = 0 x₂ = -3 x₃ = 3

y₁ = 4         y₂ = 1 y₃ = 1

The following pictorial representation helps us to calculate the area of the triangle easily.

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Area of △BCA = ½ {(x₁y₂ + x₂y₃ + x₃y₁) –  (x₂y₁ + x₃y₂ + x₁y₃)}

Substituting the values, we get

= ½ {(0 × 1) + (-3 × 1) + ( 3 × 4) – ( -3 × 4) + (3 × 1) + ( 0 × 1)}

= ½ {(0) + (-3) + (12) – (-12) + (3) + (0)}

= ½ (9) –  (-9)

= ½ (9 + 9)

= ½ (18)

= 18/2

= 9 square units.

Hence, the area of △ABC = 9 square units.

Conclusion

For finding the area of triangles using coordinates of their vertices, we need to be thorough with the coordinate geometry formulas and their related concepts. The formulas and method of finding the area of the triangle using coordinates given in this article will help you to solve the problems related to this topic easily. Hence, it is beneficial to refer to this while solving questions based on coordinate geometry.

Maths is all about numbers. Numbers are put into two categories, one is an odd number and the other one is an even number. The whole numbers that cannot be divided into pairs are known as odd numbers. When divided by 2 the odd numbers give a reminder ‘1’.The sum of two odd numbers gives an even number. The product of two or more than two odd numbers gives an odd number. The product of an even number and the product of an odd number is always an even number. In the number line, the first odd number is 1. The integer which is not an odd number is an even number. When an odd number is divided by two the result always comes out as a fraction.

Sum of Squares of n Natural Numbers Formula

The Sum of squares is the sum of the squares of numbers. Generally, it is the addition of the squared numbers. The squared terms can be of two terms, three terms, or even of ‘n’ terms. The first n even terms or the odd terms are the set of natural numbers or the consecutive numbers, etc. This is the basic math used to perform the arithmetic operation of the addition of the squared numbers.

In arithmetic operations, we often come across the sum of ‘n’ numbers. There are many formulas as well as techniques for the calculation of the sums of squares. In the statistics. It is always equal to the sum of the squares of the variation between the individual values and also the mean that is [Sigma(X_{i} + overline{X})^{2}] 

The squares formula is always used to calculate the sum of two or more than two squares in an expression. To describe how well a model can represent the data being modeled the sum of squares formula is always used. The sum of the squares is the measure of the deviation from the mean value of the data. Therefore it is calculated as the total summation of the squares minus the mean.

The sum of the squares can be calculated using the formulas that are by the algebra and by the mean. The formula to calculate the sum of the squares of the two values are:

What is the sum of squares formula in statistics, algebra, and in ‘n’ terms?

Where,

Calculating the sum of squares of the data has many applications in real life. In statistics, the sum of squares measures how far individual measurements are from the mean. This formula is used to describe how well a model represents the data being modeled and it also gives the measure of deviation from the mean value. This is why the formula is calculated as the subtraction of the total summation of the squares and the mean. It is a very useful tool to evaluate the overall variance of a data set from the mean value. It is very useful in many situations for example, the instability in the cardiovascular system that requires medical attention. Statisticians and scientists use this tool. 

An example of the sum of squares formula

Example 1: Find the value of 42 + 62 using the sum of squares formula?

Solution:

To find: 42 + 62

Given: a = 4, b = 6 

Using the sum of squares formula,

 a2 + b2 = (a + b)2 − 2ab

42 + 62 = (4 + 6)2 − 2(4)(6)

= 100 − 2(24)

= 100 − 48

= 52

Answer: The value of 42 + 62 is 52.

 

Example 2: Sum up the following series. 12 + 22 + 32 ……. 1002

Solution:

Find the sum of the series

Using the sum of squares formula, find the sum of the series.

Sum of Squares [= 1^{2} + 2^{2} + 3^{2}……….n^{2} = frac{n(n+1)(2n+1)}{6} ]

Given: n = 100 

= [100(100+1)(2×100+1)]/6

= [100(101)(201)]/6

= 338350.

Answer: Sum of the given series is 338350.

 

Example 3: Determine the sum of the squares of 10 and 22 directly using the formula. Verify the answers.

Solution: 

102 + 222 = 100 + 484 = 584

Using the formula a2 + b2 = (a + b)2 – 2ab, we get 102 + 222 = (10 + 22)2 – 2 × 10 × 22 

= 322 – 440

= 1024 – 440 = 584, Thus verified.

The term Trigonometry has been derived from two Greek words, trigonon and metron, which means triangle and to measure respectively. However, until the 16th century, it was mainly about computing the values of the missing parts of the triangle numerically when the other values of the parts were mentioned. Keeping this in mind, let’s start with an example to make the concept more clear. The third and the remaining two angles can be calculated if only the length of the two sides of the triangle and the enclosed angle’s measure is given. This distinguishes Trigonometry from geometry as it investigates the qualitative relations. Nevertheless, Trigonometry was considered a part of geometry while it became a separate branch of math during the 16th century.

 

In a modern sense, Trigonometry started with the Greeks while the first one to make a table of the values of the Trigonometric function was Hipparchus. During his time, these things were expressed in geometric terms like relations between the angles that subtended them and the chords of various forms. Moreover, until the 17th century, the modern symbols of Trigonometric functions were not established.

 

However, during the 16th century, Trigonometry started to change its character to an algebraic-analytic subject from a geometric discipline as two developments were spurred by the Frenchmen. It was the rise of symbolic algebra and the invention of analytic geometry. 

 

Another aspect of Trigonometry that received great attention was surveying, following which the triangulation’s method was first suggested in 1533 by Gemma Frisius, a Dutch Mathematician. This method was further carried for the first time on a large scale by Willebrord Snell, a Dutchman, who surveyed a stretch of 130 km using around 33 triangles. Keeping these developments in mind, the French government undertook the responsibility to triangulate the entire country under the leadership of Jean Picard, an astronomer. Following this, the survey of the entire continent of India was taken by the British as a means to have an ambitious task. Popularly known by the name of the Great Trigonometric Survey, this survey lasted from 1800 to 1913, which led to the discovery of Mount Everest, the tallest mountain peak on Earth.

Owing to these developments, the scientists from the 18th century diverted their attention to the various aspects of the functions of Trigonometry. 

Origin of the Trigonometry

For the very first time, the idea of ‘sine’ was used in the work by Aryabhatta in A.D 500. He took the word ardha-jya, which was later shortened to jive or jya in the due course. However, when translated, the word jiva became sinus, which meant curve in Latin when translated from the Arabic version. Following this, the word sinus was used as sine in the coming era in mathematical terms throughout Europe. Besides, the first abbreviated notation, known as sin, was termed by an English Professor of astronomy, identified as Edmund Gunter.

The terms ‘tangent’ and ‘ cosine’ originated much later in the time while the need to compute the sine of the complementary angle gave birth to the function of cosine, which was called Kotijya by Aryabhatta. However, Edmund Gunter, the English Professor of astronomy, termed it Cosinus while an English Mathematician, known as Sir Jonas Moore, abbreviated it as ‘cos’ in the year 1674.

Moreover, during the middle ages, Jewish and Arab scholars also played an important role in contributing to the branch. Tangents and cotangents the first table was made by Habash al-Hasib, who also wrote on the topics like astronomy. 

Moreover, the main functions based on a right-angled triangle in Trigonometry are Sin, Tan, and Cos.

Trigonometric Ratios

There are around 6 Trigonometry ratios consisting of sine, cosine, tangent, cosecant, secant, and cotangent, which tells the different possible combinations in a right-angled triangle. 

Therefore, the use of Trigonometry in solving a side in right-angled triangles is only possible when the length of the other two sides of the length and angle of a side is already known. For that, one has to choose a ratio that contains the unknown and the given side before using algebra to find the value for the unknown side. 

We know in Trigonometry we are going to discuss the relationship between angle and length in right-angled triangles. Let’s first remind ourselves, what we call sides of the right angled triangle.

 

Observe that above the right-angled triangle ABC is right angled at C. If we consider angle∠B then side AC will be perpendicular, side BC will be called based and side AB is called hypotenuse. Similarly, If we consider angle ∠A then BC will be perpendicular, AC will be considered base and AB will hypotenuse. Among these three, different ratios will give sin, cos, tan, cot, sec, cosec ratios.

 

Sin Theta Formula

So now we’ll discuss what sin theta and sin formula are. As we have discussed above, sin is a Trigonometric ratio that is perpendicular/hypotenuse.

 

 

Here x represents the angle which is under consideration.

 

Example on Sin x Formula

1. If Cos x = 35, then find the value of Sin x.

Then Find the Value of Sin x.

Solution: We know that, cos θ = BaseHypotenuse

On comparing the given ratio, Base = 3, Hypotenuse= 5.

Now we also know Pythagoras theorem, which says,

(Hypotenuse)² = (Base)² + (Perpendicular)²

⇒ (Perpendicular)² = (Hypotenuse)² – (Base)²

⇒ (Perpendicular)² = 5² – 3²

⇒ (Perpendicular)² = 25 – 9

⇒ (Perpendicular)² = 16

⇒ (Perpendicular)² = 4

Here we are considering only positive signs because the length of the side can’t be negative.

Sin x – 45

Hence this is the required answer.

 

2. If Cosec x = 67 then Find the Value of Sin x.

 

Solution: We know that sin x = 1cosec x

Using the above value we have, sin x = 16/7

Hence, sin x = 76

 

Brief Overview of Trigonometry

As the name might give you an idea, Trigonometry is all about triangles.

 

Trigonometry is the branch of mathematics that deals with the relationship between angles and the length (not arc length) of a line made by angle. Whenever we talk about angles, we need to have 2 intersecting lines, through which angle has been made. Adding to that we need one more line on which we’ll measure the length. So, in a nutshell, we need three lines. That’s what “tri” stands for. Now there are many types of triangles we know. Which type of triangle should be? So, the triangle should be right-angled. 

 

If we look deep into it, the system of Trigonometry developed to get angles and distances computed in such fields as mapmaking, astronomy, artillery range finding, and surveying while plane Trigonometry covers problems involving distances and angles in a plane. However, in spherical Trigonometry, it becomes a three-dimensional space.

 

Nevertheless, Trigonometry is that branch of the system, which helps us determine the unknown or missing sides, lengths, and angles in a given triangle. Specifically, one of the internal angles in the triangle should be 90 degrees. Now we’ll see different types of Trigonometric ratios after taking a peek into its origin.