Learn Trigonometry Test Questions

Introduction to learn trigonometry test questions:
Here we are going to see the article as learn trigonometry test questions, generally trigonometry is used to study about the triangle especially right angle triangle ,the main purpose of trigonometry is  used to find the sides and angle of a right angle triangle with the help of a trigonometric functions  such as sin ,cos, tan ,secant ,cosecant and cot. Let us start to learn some of the trigonometry test questions.


Example 1-learn trigonometry test questions:

Suppose A and B are positive acute angles and the value of cos A = `12/15` , and sin B =` 8/10` , what is the value of Sin (A + B)?

Solution:

`cos A = 12/15` , from that we find the value of sin A

Already we know that the `cos A= (adjacent side)/("hypotenuse") `

So adjacent is 12 and hypotenuse is 15

Use the Pythagorean Theorem here to find the unknown side it will be shown in below,

`(Opposite side)^ 2= ("hypotenuse")^ 2- (adjacent)^2 `

So adjacent is 12 and hypotenuse is 15

`"= (15)^2-(12)^2=225-144=81=9^2`

Opposite side =9

So `sinA =(opposite side)/("hypotenuse") `

The value of `sin A is 9/15`

Similarly

`sin B = 8/10` , from that we find the value of sin B

Already we know that the `sin B = (opposite side)/("hypotenuse") `

So opposite side is 8 and hypotenuse is 10 now use the Pythagorean Theorem to find the adjacent values

`(Adjacent side)^2 ` `= (10)^2-(8)^2=100-64=36=6^2`

So the value of adjacent side is `6/10`

`cos B=6/10`

`"cos A = 12/15, sin B=8/10, sin A =9/15, cos B = 6/10`


sin (A+B) = sinAcosB + cosAsinB
sin (A+B)= `(9/15) xx (6/10) + (12/15) (8/10)`

sin (A+B)= `(54/150) + (96/150) =1`

Example 2-learn trigonometry test questions:

Suppose the adjacent side of a right angle triangle is 6cm and opposite side of a triangle is 8cm find the hypotenuse of a triangle?

Solution:

Here the adjacent side of a right angle triangle is given as 6cm and opposite side of a right angle triangle is given as 8cm

We know that formula for Pythagorean Theorem

`AC^2=AB^2+BC^2`

Here AC is the hypotenuse

AB is the opposite side and BC is the adjacent side of a triangle

Plug those values in the above formula means we get the hypotenuse value

`AC^2=6^2+8^2`

`AC^2=36+64=100`

`AC^2= (10)^2`

So the value of hypotenuse is 10cm

I have recently faced lot of problem while learning cbse class x sample papers, But thank to online resources of math which helped me to learn myself easily on net.

Some of the trigonometry test questions with the answer key:


                 `"sin^3A+cos^3A `
1) prove   _______________     = 1- sin A Cos A
                    sin A + cos A

2)Suppose the opposite side of a right angle triangle is 15cm and hypotenuse of a triangle is 17cm find the adjacent side of a triangle?

Answer:

8cm

Learn Independent Events

Learn Independent Events
An event is a one or more possible outcomes of a certain experiment. An event is called independent event if one event does not affect the other event. For example, choosing a 7 and 8 in the deck of card with replacement is two independent events. An event consisting of more than one simple event is called compound event. In this lesson we will learn about probability of independent events.


Learn Independent Events – Learn Example Problems


Example 1: A box contains 10 bulbs, in which 5 are red color and 5 are orange color bulbs. Two bulbs are drawn one by one with replacement of first bulb. What is the probability of drawing a red and orange bulb successively?
Solution:

Let S = Sample space, n(S) = 5 + 5 = 10

A be the event of drawing a red ball first, n(A) = 5

B be the event of drawing a orange bulb second, n(B) = 5

P(A) = `(n(A))/(n(S))` =` 5/10` = `1/2`

P(B) = `(n(B))/(n(S))` = `5/10` = `1/2`

P(A and B) = P(A) · P(B) = `1/2` · `1/2` = `1/4`

P(A and B) = `1/4` .

Example 2: A box contains 15 candies, in which 5 are lemon candies, 5 are pineapple candies, and 5 are orange candies. Three candies are drawn one by one with replacement of previous candy. What is the probability of drawing a lemon, pineapple, and orange candy successively?
Solution:

Let S = Sample space, n(S) = 5 + 5 + 5 = 15

A be the event of drawing a lemon candy first, n(A) = 5

B be the event of drawing a pineapple candy second, n(B) = 5

C be the event of drawing a orange candy third, n(C) = 5

A, B, and C are independent event, so one event does not affect the other events.

P(A) = `(n(A))/(n(S))` =` 5/15` = `1/3`

P(B) = `(n(B))/(n(S))` = `5/15` = `1/3`

P(C) = `(n(C))/(n(S))` = `5/15` = `1/3`

P(A and B and C) = P(A) · P(B) · P(C) = `1/3` · `1/3` · `1/3` = `1/27`

P(A and B and C) = `1/27` .

I have recently faced lot of problem while learning cbse class 9 syllabus, But thank to online resources of math which helped me to learn myself easily on net.

Learn Independent Events – Practice Problems


Problem 1: A box contains 25 bulbs, in which 15 are red color and 10 are orange color bulbs. Two bulbs are drawn one by one with replacement of first bulb. What is the probability of drawing a red and orange bulb successively?

Problem 2: A box contains 25 candies, in which 10 are lemon candies, 10 are pineapple candies, and 5 are orange candies. Three candies are drawn one by one with replacement of previous candy. What is the probability of drawing a lemon, pineapple, and orange candy successively?

Answer: 1) ` 6/25` 2) `4/125`

Learn Intercept Group

Introduction for learn intercept group:

The intercept group has different types, one is x intercept and second one is y intercept and last one is z intercept. X intercept means point crosses the x axis and y intercept means point crosses the y axis of the line. When x = 0 and y= 0 that intercept is called as z intercept. Let us learn about the intercept group example problems and practice problems are given below.

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Learn intercept group example problems:


Example 1: Find the x intercept of the line, 2.5x + 5y = 20

Solution:

The slope intercept form is y = mx + b, where m is the slope of the line.

Given equation is in the form of ax + by = c. To find the x- intercept Plug y = 0 in the equation

Here x intercept, so y = 0

2.5x + 5(0) =20

2.5x = 20

Divide by 2.5 on both sides.

` (2.5x)/(2.5)` = ` (20)/(2.5)`

After simplify this, we get

x = 8

x intercept = 8

Example 2: Find the y intercept of the line, 12x + 6.5y = 45.5

Solution:

The slope intercept form is y = mx + b, where m is the slope of the line.

Given equation is in the form of ax + by = c. To find the y- intercept Plug x = 0 in the equation

Here y intercept, so x = 0

12(0) + 6.5y = 45.5

0 + 6.5y = 45.5

6.5y = 45.5

Divide by 6.5 on both sides.

`(6.5y)/(6.5)` = ` (45.5)/(6.5)`

After finding this, we get

y = 7

y intercept = 7

Example 3: Solve the z intercept of the given equation: 0.5x - 1.5y + 3.5z = 35

Solution:

The given equation is 0.5x - 1.5y + 3.5z = 35. Let us find z intercept.

Substitute x = 0 and y = 0

0.5(0) - 1.5(0) + 3.5z = 35

0 - 0 + 3.5z = 35

After simplify this, we get

3.5z = 35

Divided 3.5 on both sides, we get

`(3.5z)/(3.5)` = `(35)/(3.5)`

After simplify this, we get

z intercept = 10

These example problems are very helpful to learn of intercept group.


Learn intercept group practice problems:


Problem 1: Find the x intercept of the line, 2.5x + 1.5y = 40

Answer: x intercept = 16

Problem 2: Find the y intercept of the line, 5x + 14.5y = 58

Answer: y intercept = 4

Problem 3: Solve the z intercept of the given equation: 6x - 4y + 125z = 250

Answer: z intercept = 2

How to Learn My Numbers

Introduction for how to learn my numbers:
A numbers is a mathematical object used in counting and measuring. The notational symbols which represent a number is called a numeral, but in common usage the word number is used for both the abstract object and the symbol, as well as for word for the number. In addition to their used in counting and measuring, numerals are often used for labels (telephone numbers), for ordering (serial numbers), and for codes. In this article we shall dicuss  about how to learn my numbers. (Source.Wikipedia)


Classifications learn my numbers:


Types:

1. Natural numbers

2. Integers

3. Rational numbers

4. Real numbers

5. Complex numbers

1. Natural numbers:

The numbers 1, 2, 3… are called learn my natural numbers. They are also called counting learn numbers since they are used for counting objects.

2. Integers:

The numbers 0, 1, -1, 2, -2 … are called learn integers of which 1, 2, 3 … are called positive integers and -1, -2, -3… are called negative integers. The collection of all integers is denoted by the letter Z. Thus Z = {…, -3, -2, -1, 0, 1, 2, 3…}.

3. Rational numbers:

A number of the form t/r where t and r are integers and r ? 0 is called a rational number. The collection of all my rational numbers is denoted by Q. A rational number t/r is said to be in the proper form if r is a positive integer and t and r have no common factor other than 1.

Example, rational numbers `7/13` ,` 2/7`

Real numbers:

Combination for rational and irrational numbers are called as real numbers

Example: 42, -55/98

Complex numbers:

Learn complex number is of the form s + it where ‘s’ and ‘t’ are real numbers and i is called the imaginary unit, having the property that i2 = - 1. If z = s+ it then s  is called the real part of z, denoted by Re(z) and t is called the imaginary part of z and is denoted by Im(z). Examples forf complex numbers are 9 - i2,


Examples for learn my numbers:


Example 1:

Determine the rational number represented by 77.0.

Solution:

Let x =77.0. Then x = 0.777777…

? 100x = 77.777777…

? 100x - x = (77.777777…) - (0.777777…) = 77.0000…

? 99x = 77 or x = `77/99` =`7/9`

Example 2:

Write the real and imaginary parts of the following numbers:

(i) 6 - i 3 (ii)`93/2` i

Solution:

(i) Let z = 6 - i 3 ; Re(z) = 6, Im(z) = - 3

(ii) Let z =`93/2` i ; Re(z) = 0, Im(z) =`93/2`

Learn Roman Numerals Properly

Introduction to Roman numerals:

Roman numerals are a numerals system of ancient Rome based on letters of the alphabet, which are combined to signify the sum (or in some cases, the difference) of their numbers. The first ten Roman numerals are as follows:

I, II, III, IV, V, VI, VII, VIII, IX, and X.

In this be article we are going to learn Roman numerals properly. (Source: Wikipedia)


Learn Roman numerals properly:


All Roman numerals should be formed by combining the symbols that the Romans used.

That is I , V , X , L , and C are the letters used as Roman numerals.

Learning rules of Roman numerals properly:

In order to learn Roman numerals properly the best way is to start from the rules.

Rule 1:

Add that number if one or more letters are placed subsequent to another letter of greater number.

Examples are:

III = 1+1+1 = 3

VIII = 5+1+1+1 = 8

Rule 2:

Subtract that number if a letter is placed previous to another letter of greater number.

Examples are:

IV = 4 (5-1 = 4)

XC = 90 (100-10 =90)

XLI = 41 (50-10+1 = 41)

Rule 3:

We should only subtract exponents of 10 (I = 100, X = 101, C = 102)

Examples are:

IX = 9 (10-1 = 9)

XL = 40 (50-10 =40)

CD = 400 (500-100 = 400)

Rule 4:

We shouldn’t subtract more than 1 number from another number.

XXXVIII = 38 (10+10+10 + 5 + 1 + 1 + 1 = 18)

We shouldn’t write 38 as IIXXXX.


Learning example problems for Roman numerals properly:


Learning problem 1:

Write 7 in decimal number.

Solution:

The given decimal number can be written as 7 = 5+1+1

Note that the numbers given must be splitted only with the fundamental numerals I , V , X , L , C , D , M

Therefore 7 = 5+1+1

V  I  I

Where V indicates decimal number 5

I indicate decimal number 1

In roman numerals 7 can be written as VII.

Learning problem 2:

Write 19 in decimal number

Solution:

The given decimal number can be written as 19 = 10+(10-1)

Therefore 19 = 10+(10-1)

Where X indicates decimal number 10

I indicate decimal number 1

In roman numerals 19 can be written as XIX.

Learning problem 3:

Write 44 in decimal number

Solution:

The given decimal number can be written as 44 = (50-10)+(5-1)

Therefore 44 = 50+(5-1)

Where L indicates decimal number 50

X indicates decimal number 10

V indicates decimal number 5

I indicate decimal number 1

In roman numerals 44 can be written as XLIV.

How to Learn Multiplication Fast

Introduction to how to learn multiplication fast:
Multiplication is one of the basic operations in mathematics. Multiplication can be done using math tables. Multiplying a small number is a simple task, using math table, but when we consider for a larger number, this process is tedious. We have introduced a lot of easy ways for multiplication to learn as fast as possible. Here we are going to see how to learn multiplication fastly.


Methods to learn multiplication fast:


Many new trends in mathematics are introduced how to learn multiplication fast. These fast methods save our computational processing time during multiplication of complex problems.

Here we use some of the fast ways to learn how to do multiplication effectively,

Methods to learn multiplication fast for two digit number less than 20:

While multiplying any two digit number that is less than 20, follow the following steps,

Let us consider, 12 * 14

Choose the largest number in the front.
Here the largest number is 14. Hence the sum becomes 14 * 12.
Add the largest number (14) with the last digit of the lowest number (2).
Hence it becomes 14 + 2 = 16.
Multiply 10 with the resulting answer.
Hence we get 16 * 10 = 160.
Now multiply the last terms of two given numbers.
Hence we get, 4 * 2 =8.
Add 160 + 8 = 168.
Thus we got the right answer as 168.

Thus we learn how to multiply fastly for two digit number less than 20.

Another example:

Let us consider, 15 * 17

Choose the largest number in the front.
Here the largest number is 17. Hence the sum becomes 17 * 15.
Add the largest number (17) with the last digit of the lowest number (5).
Hence it becomes 17 + 5 = 22.
Multiply 10 with the resulting answer.
Hence we get 22 * 10 = 220.
Now multiply the last terms of two given numbers.
Hence we get, 7 * 5 =35.
Add 220 + 35 = 255.
Thus we got the right answer as 255.


Methods to learn multiplication fast for any number with multiples of 9:

While multiplying for any number with multiples of 9, follow the following steps,

Let us consider, 64 *99

Take 64, add two zeros with 64, because we have two 9. (If three 9, add 3 zeros).
Hence it becomes 6400.
Now subtract 6400 with the original number 64.
Hence we get, 6400 – 64 = 6336.
Thus we got the right answer as 6336 in a fast way.

Another example:

Let us consider, 34 *99

Take 64, add two zeros with 34, because we have two 9. (If three 9, add 3 zeros).
Hence it becomes 3400.
Now subtract 3400 with the original number 34.
Hence we get , 3400 – 34 = 3366.
Thus we got the right answer as 3366 in a fast way.

Thus we learn how to multiply fastly any number with multiples of 9.

Learn How to Count Percentage

Introduction to learn how to count percentage

In mathematics, a percentage is a way of expressing a number as a fraction of 100 (per cent meaning "per hundred" in French). It is often denoted using the percent sign, "%", or the abbreviation "pct". For example, 45% (read as "forty-five percent") is equal to 45 / 100, or 0.45. Counting percentage means the process of expressing the value of a number in 100. (Source: From Wikipedia).

Here we are going to learn how to count percentage.

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Example problems to learn counting percentage

Here we will learn some example problems to count percentage.

Example 1

Write the fraction `30/100` as a percentage

Solution

Here 30 is expressed as a fraction of 100. To write `30/100` as a percent we have to multiply `30/100` by 100.

We, get `30/100` * 100 = 30

So, 30/100 is equal to 30 percentage or 30%

Example 2

Count the percent of 30 in 80

Solution

Here we have to count the percent of 30 in 80.

Let x be the percentage,

So, `30/80` = `x/100`

x = `30/80` * 100

x = `300/8`

x = 37.5

So, 30 is 37.5 percent of 80

Example 3

What is 30 percent of 20.

Solution

Here we have to find the value of 30 percent in 20

Let the number be x

So, `x/20` = `30/100`

x = `30/100` * 20

x = `600/100`

x = 6

So, 30 percent of 20 is 6.


Few more examples to learn counting percentage

Here we will learn few more problems to count percentage.

Example 1: Write the decimal 0.25 as fraction

Solution

0.25 can be written as 25/100 in fraction

To convert this fraction into percentage we have to multiply the fraction by 100

Doing so, `25/100` * 100 = 25

So, 0.25 is 25 percent.

Example 2: Write the proportion 32:100 as a percent

Solution

The proportion 32:100 can be written as `32/100` in fractional form.

Now, `32/100` can be written as `32/100` * 100 = 32 percent

So, 32:100 is equal to 32 percent.

Example 3: 25 is 4 percent of what number.

Solution

Let the number be x

So, `25/x` = `4/100`

This equation can be written as, x = `25/4` * 100

x = 25 * 25

x = 625

So, 25 is 4 percent of 625.