Intersecting Acute Angles

The intersecting acute angles measure the 0° and 90°. A sketch of angles is the main topic in geometry. Two lines join at a single point to form an angle.

If P and R are two straight lines they intersect at a point Q and make an angle Q. The angle Q is also represented as PQR. Here center point represents the vertex of the angle.

Example problems for intersecting acute angles:

Problem 1:-

Solving the intersecting acute angle from the where slopes  m1= 4 and m2=14

Solution:

In above diagram show the acute angles intersecting the two lines

tan phi = (m2- m1)/ (1+m1m2)

= (14-4)/(1+(14x4))

=10/ 57

= 0.175

Hence we find the phi value.

phi = arc tan(0.175)

from the tangent table,we get
phi =1o  (rounded value)



from the given example we can find the obtuse angle.

To subtract the acute angle between the the two values L,L2 from the 1800 straight angle.

That is,

phi = 180- phi

phi = 180 -1o

phi = 179o

Hence we find the obtuse angle.

Problem 2:-

Solving the intersecting acute angle from the where slopes  m1= 20 and m2=28

Solution:

In above diagram show the acute angles intersecting the two lines

tan phi = (m2- m1)/(1+m1m2)

= (28-20)/(1+(28x20)

= 8/ 561

= 0.0142

Hence we find the phi value.

phi   = arc tan(0.0142)

from the tangent table,we get
phi   = 1o  (rounded value)

from the given example we can find the obtuse angle.

the obtuse angle is calculated by subtracting the acute angle between the value of L and L2 from the straight angle 180 0

ie phi = 180- phi

phi = 180 -1o

phi = 179o

Hence we find the obtuse angle.

My forthcoming post is on algebra 2 formulas sheet and sample papers for class 10 cbse will give you more understanding about Algebra.

Practice problems for intersecting acute angles:

1. Solving the intersecting acute angle from the where slopes  m1= 2 and m2=4

Answer:- 168o

2. Solving the intersecting acute angle from the where slopes  m1= 4 and m2= 8

Answer:- 173o

3. Solving the intersecting acute angle from the where slopes  m1= 8 and m2=12

Answer:- 178 o

4. Solving the intersecting acute angle from the where slopes  m1= 12 and m2=14

Answer:- 174 o

Time Value of Money

Time value of money is one of the most important concepts in mathematics, also widely used in real life finance and economical scenarios. Time value of money is taught sometime in high school mathematics that plays a major role in the practical life. Let’s discuss about the same in this post.

What is Time Value of Money or TVM?
Time value of money is popularly abbreviated as TVM. The concept of Time value of money defines that the value of money keeps changing with time. Most of the time, the value of money today is lesser than the value of money tomorrow. For example: The popular parenting magazine cost was Rs.50 till last year. Today, the cost of the same parenting magazine is Rs.75. This demonstrates that in the period of one year, the value of money has changed. Understanding the concept of Time value of money requires understanding in related concepts like Present value and Future value.Present Value and Example:

The current worth of money is termed as the present value. For example: Only for today they have offered the scheme of Rs.699 for post pregnancy weight loss program. From tomorrow the rate of post pregnancy weight loss program will be again Rs.1799. Therefore, the present value here is Rs.699. Based on the present value and the difference in future, one can find out the rate of increase or decrease in TVM.

Future Value and Example:
Future value is the amount or value of money on a specified date in future with respect to the same in today’s date. For example: The book on develop reading habit cost is Rs.100 today but the cost of same book on develop read habit will increase to Rs.150 by next month, owing to its popularity. Thus, Rs.150 is the future value in this situation.
These are the basics on Time Value of Money.

Coordinate Plane

What is a Coordinate Plane?

The coordinate plane is a two-dimensional surface on which we can plot points, lines and curves. It has two scales, called the x-axis and y-axis, at right angles to each other. The plural of axis is 'axes' (pronounced "AXE-ease").


Definitions

X axis

The horizontal scale is called the x-axis and is usually drawn with the zero point in the middle. As you go to the right on the scale the values get larger and are positive. As you go to the left, they get larger but are negative.
Y axis

The vertical scale is called the y-axis and is also usually drawn with the zero point in the middle. As you go up from zero the numbers are increasing in a positive direction. As you go down from zero they increase but are negative.


Origin

The point where the two axes cross (at zero on both scales) is called the origin. In the figure above you can drag the origin point to reposition it to a more suitable location at any time.


Quadrants



Quadrants of the coordinate plane When the origin is in the center of the plane, they divide it into four areas called quadrants. The first quadrant, by convention, is the top right, and then they go around counter-clockwise. It is conventional to label them with numerals but we talk about them as "first, second, third, and fourth quadrant".

Coordinates of a point

The coordinates of a point are a pair of numbers that define its exact location on a two-dimensional plane. Recall that the coordinate plane has two axes at right angles to each other, called the x and y axis. The coordinates of a given point represent how far along each axis the point is located.
Ordered Pair

The two numbers in parentheses are the x and y coordinate of the point. The first number (x) specifies how far along the x (horizontal) axis the point is. The second is the y coordinate and specifies how far up or down the y axis to go. It is called an ordered pair because the order of the two numbers matters - the first is always the x (horizontal) coordinate.

The sign of the coordinate is important. A positive number means to go to the right (x) or up(y). Negative numbers mean to go left (x) or down (y). (The figure at the top of the page has the values of the axes labelled with the appropriate sign).
Abscissa

The abscissa is another name for the x (horizontal) coordinate of a point. Pronounced "ab-SISS-ah" (the 'c;' is silent). Not used very much. Most commonly, the term "x-coordinate" is used.
Ordinate



The ordinate is another name for the y (vertical) coordinate of a point. Pronounced "ORD-inet". Not used very much. Most commonly, the term "y-coordinate" is used.

Pan Balance Math Problems

A balance is a beam balance or laboratory balance or pan balance used to measure the accurate value of mass of an object. A weighting pan and scale pan span is hold above the machine. Pan balance is used to distribute weight between two containers in a machine balance.  By using pan balance we have to calculate the weight of the objects and evenly distribute the objects. Let us see about pan balance math problems in this article.


Worked Examples to Pan Balance Math Problems

Example 1 to Pan Balance Math Problems:

Solve the pan balance math problem.

Pan Balance Math Problems

Find the weight of the two rectangles in the above problem?

Solution:

Step 1:

Given weight in left side object = 6 kg

Given weight in the right side object = 11 kg

Step 2:

In left side, the pan is not balanced if we remove the two rectangles.

So, subtract the right side object weight – left side present object weight, we will get the result.

Step 3:

Subtract the amount on left side object from the right side object, we get,

11 – 6 = 5 kg

Step 4:

Two rectangles have the total of 5 kg.

Step 5:

Two rectangles in the left side of the pan balance are of the same size.

Therefore, each rectangle is of 2.5 kg.

Example 2 to Pan Balance Math Problems:

Solve the pan balance math problem.

Pan Balance Math Problems

Find the weight of the diamond object in the above problem?

Solution:

Step 1:

Total weight in the right side of the pan balance = 17 kg

Triangle object = 4 kg

Square object = 8 kg

Diamond object =?

Step 2:

In left side, the pan is not balanced if we remove the diamond object.

So, subtract the right side object weight – left side present object weight, we will get the result.

Step 3:

Adding the triangle and square shaped object weight = 4 + 8 = 12

Left side = 12 kg

Step 4:

Subtracting the right side objects weight – left side objects weight = 17 – 12 = 5 kg

Step 5:

Therefore, the weight of the diamond object is 5 kg.



Practice Problems to Pan Balance Math Problems

Solve the pan balance math problem.

Pan Balance Math Problems

Find the weight of the diamond object in the above problem?

Solution:

1. Pan balance weight in the right side is 10 kg.

Proving Identities Calculator

An identity is a relative which is tautologically factual. These are typically taken to denote something that is factual by meaning, either straight by the definition, or as a result of it. For model, algebraically, this occurs if an equation is content for all values of the concerned variables. Definitions are frequently indicated by the 'triple bar' sign ≡, such as A2 ≡ x·x. A calculator is a little (often small), typically cheap electronic machine used to do the basic operations of mathematics. Present calculators are additional moveable than most computers, although most PDAs are similar in size to handheld calculators.


Proving Algebra and Trigonometric Identities Calculator:

proving identities calculator

c + 0 = c (additive identity)
c + (-c) =0 (additive inverse)
c x 1 = c (the multiplicative identity)
c  (1/c )= 1 (the multiplicative inverse)
sin2a≡ 1 – cos2a
cos2a≡ 1 – sin2a
tan2a ≡ sec2a − 1

Examples of Proving Algebra Identities:

Study example 1:

21+ 0 = 21

Solution:

(We know that the additive identity in algebra (c+0 = c). Therefore the answer is 21)

Study example 2:

23+(-23) = 0

Solution:

(We know that the additive inverse in algebra (c+(-c) = 0) this problem answer 0.)

Study example 3:

1x14= 14

Solution:

(We know that the multiplicative identity  (1x c = c). So this problem answer 14.)

study example 4:

23 (1/23) = 1

Solution:

(We know that the multiplicative inverse  (c x 1/c = 1). So this problem answer is 23.)

Between, if you have problem on these topics math 2nd grade word problems, please browse expert math related websites for more help on how to prepare for iit jee 2013.

Examples for Proving Trigonometric Identities:

Study example 1:

Prove the following identity: (sin M)4 +2(sin M)2(cos M)2 + (cos M)4 = tan M cot M

Solution:

(sin M)4 +2(sin M)2(cos M)2 + (cos M)4 = tan M cot M

(sin M)4 +2(sin M)2(cos M)2 + (cos M)4 =(sin M/cos M)xx(cos M/sinM)

(sin M)4 +2(sin M)2(cos M)2 + (cos M)4 = 1
((sin2M)2 + (cos2M)2)((sin2M)2 + (cos2M)2) = 1       { (a+b)2=a2+b2+2ab}

( (Sin2M)+cos2M )2 =1

(1)(1) = 1

Study example 2:

Prove the following identity: tan(p)cos(p) = sin(p)

Solution:

tan(p)cos(p) = sin(p)

(sinP/cosP) cos p= sin(p)

sin(p) = sin(p)

Logic Problems With Solutions

logic problems with solutions set 1

Problem: - If in the code, ALTERED is written as ZOGVIVW, then in the same code, RELATED would be written as?

Solution: - Clearly, each letter of the ALTERED is replaced by the letter which occupies the same position from the other end of the English alphabet, to obtain the code. Thus, A, the first letter of the alphabet, is replaced by Z, the last letter. L, the 12th letter from the beginning of the alphabet, is replaced by O, the 12th letter from the end. T, the 7th letter from the end of the alphabet is replaced by G, the 7th letter from the beginning of the alphabet and so on.

Similarly, in the word RELATED, R will be coded as I, E as V, L as O, A as Z, T as G and D as W. Thus, the code becomes IVOZGVW.                            (Answer)

Problem: - Pointing to a photograph, a lady tells X, “I am the only daughter of this lady and her son is your maternal uncle.” How the speaker is related to X’s father?

Solution: - Clearly, the speaker’s brother is X’s maternal uncle. So, the speaker is X’s mother or his father’s wife.

Problem: - In a school, there were five teachers. A and B were teaching Hindi and English. C and B were teaching English and Geography. D and A were teaching Mathematics and Hindi. E and B were teaching History and French. Answer the following questions:

1)      Who among the teachers was teaching maximum number of subjects?

2)      Who among the teachers, teaches Hindi?

3)      D, B and A were teaching which subjects?

Solution: - The given information can be analysed as under:

Prob 1


Clearly, B teaches maximum number of subjects, i.e. 5.

Three teachers were teaching Hindi – A, B and D.

D, B and A were teaching Hindi.
Logic Problems with Solutions Set 2

Problem: - A man walks 1 km towards East and then he turns to South and walks 5 km. Again he turns to East and walks 2 km, after this he turns to North and walks 9 km. Now, how far is he from his starting point?

Solution: - The movements of the man are as shown in the figure. (A to B, B to C, C to D, D to E).

Prob2

Clearly, DF = BC = 5 km.

EF = (DE – DF) = (9-5) km = 4km

BF = CD = 2 km

AF = AB + BF = AB + CD = (1 + 2) km = 3 km.

Therefore man’s distance from starting point A

= AE = `sqrt(AF^2 +EF^2)`  = `sqrt(3^2 + 4^2)`

= `sqrt25`   =   5 km                                      (Answer)

I am planning to write more post on Rational Inequalities and ibps exams 2013. Keep checking my blog.

Problem: - Which of the following diagrams correctly represents the relationship among the classes: Tennis fans, Cricket players, Students?

Prob2

Solution: - Some students can be cricket players. Some cricket players can be tennis fans.

Some students can be tennis fans. So, the given items are partly related to each other. So the correct diagram which represents the relationship between the above is option (A).

Loci and Concurrency Theorems Tutorial

In this tutorial, we will study about definition of loci and concurrency, loci and concurrency theorems.

Definition of Loci:

If a point moves in some way which satisfies some given geometrical condition at every instant during its motion, then the path traced out by the moving point is called loci of a point.

Every point satisfies the given geometrical condition is a point of loci, and

Every point of loci should satisfy the given geometrical condition.

Singular form of loci is called as locus.

Definition of concurrency:

If three or more than three lines pass through a point, then the common point is called point of concurrency.

Let us see about loci and concurrency theorems in this tutorial.
Tutorial - Loci Theorem:

Definition:

The loci of a point which is equidistant from the two given points, is the perpendicular bisector of the line segment and joining the two given points.

Given:

There are two given fixed points M and N.

A is a moving point that is A is loci of two given points M and N.

AM = AN.

To Prove:

The loci of A is the perpendicular bisector of MN line segment.

Construction:

Triangle-Loci Theorem

Join MN. Bisect MN at B. Join AM, AN and AB.

Proof:

Therefore, A moves such that AM = AN.

As we know that,

A must pass through the point B of the line segment MN, because BM = BN.

Thus A passes through the mid-point of MN.

Now, in ∆MAB and ∆NAB,

Given    AM = AN

By construction, BM = BN

AB = AB

By sss congruence rule,

∆MAB = ∆NAB

angle MBA = angle NBA       (1)

But angle MBA + angle NBA = 180°  (2)

Substitute (1) into (2),

2 angle MBA = 180°

Divide by 2 each side.

angle MBA = 90°

Therefore, AB _|_ MN

Thus, AB is proved as perpendicular bisector of MN.
Tutorial - Concurrency Theorem:

Definition:

A concurrency point of the perpendicular bisectors of sides of a triangle is called circum-center of the triangle.

A concurrency point of the median of a triangle is called centroid of the triangle.

The concurrency diagram is shown in triangle figure beliow.

Triangle-Concurrency Theorem

A,B,C are mid-points of PQ, QR and QR respectively. PB, QC and RA are medians of ∆PQR. Here O is Concurrency point.

As we know from congruence rule,

OB _|_ QR

OA _|_ PQ

OC _|_ PR

These are all perpendicular bisectors of triangle. Therefore, O is concurrency point.