Show that : cos(36 - A) cos(36 + A) + cos(54 + A) cos(54 - A) = cos2A

Show that : 

cos(36 - A) cos(36 + A) + cos(54 + A) cos(54 - A) = cos2A

Sol :

L.H.S. = cos(36 - A) cos(36 + A) + cos(54 + A) cos(54 - A)

           = 2/2 cos(36 - A) cos(36 + A) + 2/2 cos(54 + A) cos(54 - A)
       
          = 1/2{cos(36 -A +36 +A) + cos(36-A-36-A)} + 1/2{cos(54+A+54-A) + cos(54+A-54+A)}

         = 1/2 {cos72 + cos(-2A)} + 1/2 {cos108 + cos2A}

        = 1/2 {cos72 + cos108 +2cos2A}
   
        = 1/2 {2cos (72+108)/2 .cos(72-108)/2  + 2cos2A}

        = 1/2 {2 cos90 . cos16  + 2cos2A}

        = 1/2 . 2 cos2A            [cos90 = 0]

        = cos2A

        = R.H.S         PROVED




KEYWORDS  : CBSE, Geeks Networks, NCERT solved answers, solved answers of mathematics of 11th class, trigonometric functions, trigonometry, 

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The mean and variance of 5 observations are 15 and 6 respectively.If three observations are 11,15,18, find the remaining observations.

The mean and variance of 5 observations are 15 and 6 respectively.If three observations are 11,15,18, find the remaining observations.

Sol :

Keywords : statistics,CBSE, Geeks Networks, NCERT solved answers, solved answers of mathematics of 11th class,

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profiles of users in computer system

Simply, Profile is used for desktop related settings.
There are three types of profile :-

1. Local Profile 
2. Roaming Profile
3. Mandatory Profile

1. Local Profile :- It is automatically provided by Operating System. When we create a user , he gets local profile by default.

2. Roaming Profile :- It is manually created by administrator of Domain Controller (DC).With this profile , If user will change anything in his desktop,he will get that changes in entire domain. 

3. Mandatory Profile :- This Profile is also manually created by Network Administrator.The owner/user of this profile changes desktop settings, those settings will not work longer.


How to check Profile


Select my computer ----> Right click--------> Property-------> Advance------------> User profile Setting. 

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In a parallelogram, The lengths of adjacent sides are 15 cm and 17 cm respectively. If the length of one diagonal is 18 cm, find the length of the other diagonal........

In a parallelogram, The lengths of adjacent sides are 15 cm and 17 cm respectively. If the length of one diagonal is 18 cm, find the length of  the other diagonal........

Sol :  

In a parallelogram, the sum of the squares of the diagonals = 2 x ( the sum of the squares of the two adjacent sides).

Or          (D1)^2 + (D2)^2 = 2{(a)^2 + (b)^2}

Or          18^2 + (D2)^2   = 2 { 15^2 + 17^2 }

Or          324 + (D2)^2    = 2 { 225 + 289}

Or             (D2)^2  =  1028 - 324

Or             D2       = sqrt(704)

Or             D2       =  26.53 cm

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Find the no. of Diagonals of a Polygon of 15 sides....

Find the no. of Diagonals of a Polygon of 15 sides....


Sol  :

A polygon of n sides will have n vertices . A diagonal or a side of the polygon will be formed by joining any two vertices of the polygon..

No. of Diagonals of the polygon + No. of sides of the polygon = C(n,2) - n

                                                                             = n! / 2!(n-2)!  - n

                                                                             = n x (n-1) x (n-2)! / 2!(n-2)!  - n

                                                                             = n x (n-1) / 2   - n


                                                                             = {n(n-1) - 2m} / 2

                                                                             = n{n-1-2} / 2

                                                                             = n(n-3) / 2


Now,     Putting n = 15,  We get the number of Diagonals   = 15(15-3)  / 2

                                                                              = 15 x 12 / 2

                                                                              = 90

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Given 7 flags of different colours, how many different signals can be generated if each signal requires the use of two flags, One below the other ?

Given 7 flags of different colours, how many different signals can be generated if each signal requires the use of two flags, One below the other ?

Sol  :

First place can be filled up in seven ways. After filling first place , we have left with 6 flags.
Therefore   The second place  can be filled up in six ways...

The total number of signals which can be generated = 7 x 6 ways = 42 ways..........

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How many 3-letter code words are possible using the first 10 letters of the English alphabet, if no letter can be repeated ?

How many 3-letter code words are possible using the first 10 letters of the English alphabet, if no letter can be repeated ?

Sol :

First letter of the code word can be selected in 10 ways. After the selection of the first letter, we have nine letters. Hence, the second letter can be selected  in nine ways. According to Fundamental principal of Counting(FPC), first two letters can be selected in 10 x 9 ways.

Similarly third letter can be selected in eight ways . According to FPC , number of ways of selecting Three letters out of ten letters of English alphabet = 10 x 9 x 8  ways = 720 ways...........

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How many 3-digit even numbers can be formed from the digits 1,2,3,4,5 and 6 if the digits can be repeated?

How many 3-digit even numbers can be formed from the digits 1,2,3,4,5 and 6 if the digits can be repeated?

Sol : 

Let 2 be fixed at unit's place . The ten's place can be filled up in six ways. The hundred's place can also be filled in six ways.

No. of numbers that can be formed when 2 is at unit's place  =  6x6 = 36.

Similarly when 4 is at unit's place, the number of numbers that can be formed = 36.

Again when 6 is at unit's place, The number of numbers that can be formed = 36.

Therefore   The total numbers of ways when 3-digits numbers can be formed, the digits being repeated = 36 x 3 = 108......

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How many 3-digit numbers can be formed from the digits 1,2,3,4 and 5 when repetition of digits is allowed and if repitition of digits is not allowed.......

How many 3-digit numbers can be formed from the digits 1,2,3,4 and 5 when repetition of digits is allowed and if repetition of digits is not allowed.......

Sol :

When Repetition is allowed   :

There are five digits, 1,2,3,4 and 5. Every digit can be selected any number of times. Hence , we can select first digit five times. The second digit 5 times and the third digit five times.
Therefore the number of ways in which the selection of three digits can be made  5x5x5x ways = 125 ways.

When repetition is not allowed   :

Under the condition , first digit can be selected 5 ways. After the selection of first digit four digits are left. Second digit can be selected in four ways and third digit can be selected in three ways.

Therefore  The total  number of ways of selecting three digits is 5x4x3 = 60 ways.......

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Reduce the Equation 5y+3 = 0 into intercept form and find its intercept on the axes.................

Reduce the Equation 5y+3 = 0 into intercept form and  find its intercept on the axes.................

Sol  : 



We have                        5y + 3 = 0

Or                                           5y = -3


Or                                       5y/3 = -3/3

Or                                    y/3/5   =  -1

Or                                    -y/3/5 = 1



Therefore      Intercepts on the axes are 0 and -3/5.............



Keywords  :  Geeks Networks, coordinate geometry, mathsmania, NCERT solved answers, slope of line, solved answers of mathematics of 11th class, Straight Lines,

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Reduce the Equation 3x+2y-12=0 into intercept form and find its intercepts on the axes...........

Reduce the Equation 3x+2y-12=0 into intercept form and find its intercepts on the axes...........

Sol  :

here                                 3x+2y-12=0

Or                                          3x+2y = 12

Dividing by 12    we get,

                                               x/4 + y/6 = 1

Therefore                         Intercepts on the Axes are  4 and 6............


Keywords   :   coordinate geometry, Geeks Networks, NCERT solved answers, mathsmania, solved answers of mathematics of 11th class, slope of line, Straight Lines,

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Reduce the Equation 6x+2y-5=0 into slope-intercept form and find its slope and intercept........

Reduce the Equation 6x+2y-5=0 into slope-intercept form and find its slope and intercept........

Sol   :

            Here Equation is
                                      6x+2y-5=0

Or                                           2y = -6x+5

Or                                             y = -6/2x+5/2

Or                                             y = -3x+5/2

Therefore                  Slope = -3      and Intercept = 5/2





Keywords   :   coordinate geometry, Geeks Networks, mathsmania, NCERT solved answers, solved answers of mathematics of 11th class, Straight Lines,

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Reduce the equation x+7y = 0 into slope intercept form and find its slope and the intercept.....

Reduce the equation x+7y = 0 into slope intercept form and find its slope and the intercept.....

Sol : 

             x + 7y = 0

Or        7y = -x

Or          y = -1/7x + 0

Therefore         Slope = -1/7      and     Intercept = 0







Keywords   :   CBSE, coordinate geometry, Geeks Networks, NCERT solved answers, solved answers of mathematics of 11th class, mathsmania, Straight Lines,

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