Probabiliy problem

Find the probability of getting a head when a coin is tossed once. Also find the probability of getting a tail.

In the experiment of tossing a coin once, the number of possible outcomes

is two — Head (H) and Tail (T). Let E be the event ‘getting a head’. The number of

Outcomes favorable to E, (i.e., of getting a head) is 1. Therefore,

P (E) = P (head) = Number of outcomes favorable to E = 1/2

Number of all possible outcomes

Similarly, if F is the event ‘getting a tail’, then P(F) = P(tail) =1/2

to understand more about Probabilities these website links may help you,
Probability explainations, and conditional probability

The origin of the term "Sine" in Mathematics

The origin of the term "Sine" Mathematics:
The first use of the idea of ‘sine’ in the way we use it today was in the work Aryabhatiyam
by Aryabhatta, in A.D. 500. Aryabhatta used the word ardha-jya for the half-chord, which was shortened to jya or jiva in due course. When the Aryabhatiyamm 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’.

Theorem of Triangle

Theorem of Triangle:

If in two triangles, corresponding angles are equal, then their
corresponding sides are in the same ratio (or proportion) and hence the two
triangles are similar.

Theorem of Triangle is explained below to understand it better:

This criterion is referred to as the AAA (Angle–Angle–Angle) criterion of similarity of two triangles. This theorem can be proved by taking two triangles ABC and DEF such that
∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F (see Fig. 6.24)

Cut DP = AB and DQ = AC and join PQ.

So, Δ ABC ≅ Δ DPQ (Why ?)
This gives ∠ B = ∠ P = ∠ E and PQ || EF (How?)
Therefore,
( / notation used to represent divided by) *

DP/PE =DQ/QF

AB/DE =AC/DF
Similarly,
AB/DE=BC/EF and also AB/DE=BC/EF=AC/DF

Remark : If two angles of a triangle are respectively equal to two angles of another triangle, then by the angle sum property of a triangle their third angles will also be equal.

Problem related to Arithmetic Progression

Which term of the Arithmetic Progression : 21, 18, 15, . . . is – 81? Also, is any term 0? Give
reason for your answer.

Here is the step by step explanation to the A.P problem,

Solution : Here, a = 21, d = 18 – 21 = – 3 and an = – 81, and we have to find n.

As an = a + ( n – 1) d,

we have – 81 = 21 + (n – 1)(– 3)

– 81 = 24 – 3n

– 105 = – 3n

So, n = 35

Therefore, the 35th term of the given AP is – 81.

Next, we want to know if there is any n for which an = 0. If such an n is there, then

21 + (n – 1) (–3) = 0,

i.e., 3(n – 1) = 21

i.e., n = 8

So, the eighth term is 0.

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www.tutorvista.com ,tutor.com, other links

Algebraic Method of Solving a Pair of Linear Equation

Algebraic Method of Solving a Pair of Linear Equations:

Substitution Method : We shall explain the method of substitution by taking
some examples.

Example 1 :
Solve the following pair of equations by substitution method:
7x – 15y = 2 (1)
x + 2y = 3 (2)
Solution :
Step 1 : We pick either of the equations and write one variable in terms of the other.
Let us consider the Equation (2) :
x + 2y = 3
and write it as x = 3 – 2y (3)

Step 2 : Substitute the value of x in Equation (1). We get
7(3 – 2y) – 15y = 2
i.e., 21 – 14y – 15y = 2
i.e., – 29y = –19
Therefore, y =19/29
Step 3 : Substituting this value of y in Equation (3), we get
x = 3 – 2 (19/29) =49/29

Therefore, the solution is x =49/29 , Y =19/29

Verification : Substituting x =49/29 and Y=19/29

,you can verify that both the Equations
(1) and (2) are satisfied.

Step 1 : Find the value of one variable, say y in terms of the other variable, i.e., x from
either equation, whichever is convenient.
Step 2 : Substitute this value of y in the other equation, and reduce it to an equation in
one variable, i.e., in terms of x, which can be solved. Sometimes, as in Examples 9 and
10 below, you can get statements with no variable. If this statement is true, you can
conclude that the pair of linear equations has infinitely many solutions. If the statement
is false, then the pair of linear equations is inconsistent.
Step 3 : Substitute the value of x (or y) obtained in Step 2 in the equation used in
Step 1 to obtain the value of the other variable.

Quadratic Equations

Quadratic equations:

Quadratic Equations come up when we deal with
many real-life situations. For instance, suppose a
charity trust decides to build a prayer hall having
a carpet area of 300 square meters with its length
one meter more than twice its breadth. What
should be the length and breadth of the hall?
Suppose the breadth of the hall is x meters. Then,
its length should be (2x + 1) meters. We can depict
this information pictorially as shown in Fig below


Now, area of the hall = (2x + 1). x m2 = (2x2 + x) m2
So, 2x2 + x = 300 (Given)
Therefore, 2x2 + x – 300 = 0
So, the breadth of the hall should satisfy the equation 2x2 + x – 300 = 0 which is a
quadratic equation.

Euclid’s five postulates

Euclid’s five postulates

Postulate 1 : A straight line may be drawn from any one point to any other point.

Axiom 5.1 : Given two distinct points, there is a unique line that passes through

them.

How many lines passing through P also pass through Q (see Fig. 5.4)? Only one,

that is, the line PQ. How many lines passing through Q also pass through P? Only one,

that is, the line PQ. Thus, the statement above is self-evident, and so is taken as an

axiom.

Postulate 2 : A terminated line can be produced indefinitely.

Note that what we call a line segment now-a-days is what Euclid called a terminated

line. So, according to the present day terms, the second postulate says that a line

segment can be extended on either side to form a line (see Fig. 5.5).

Postulate 3 : A circle can be drawn with any centre and any radius.

Postulate 4 : All right angles are equal to one another.

Postulate 5 : If a straight line falling on two straight lines makes the interior

angles on the same side of it taken together less than two right angles, then the

two straight lines, if produced indefinitely, meet on that side on which the sum of

angles is less than two right angles.