S3: Geometry Radian

DEGREE AND RADIAN

Both degree and radian is a measure of angle.

Radian is defined as the angle from the centre of a circle which intercepts an arc equal in length to the radius of the circle.

                  360^o = 2π rad

                  => 1^o = 2π rad / 360^o

  

        2π rad = 360^o

          =>   1 rad = 360^o / 2π

Converting between Degree and Radian

                  360^o = 2π rad

  or

                 π rad = 180^o

Example

What is the value of 240^o in rad?

     360^o = 2π rad 

                240^o = 2π rad x 240^o

                                 360^o

                        = 4π/3 rad

Example

What is the value of 3 radian in degree?

            2π rad = 360^o

                           1 rad = 360^o / 2π

                           3 rad = 360^o x 3 / 2π

                                    = 171.81^o

	Degree	Radian
	1o =  π rad       180	1 rad = 180o               π
One Circle	360	2πrad
Circumference	2πr	2πr
Arc(S)	θ    x 2πr 360	rθ
Area of circle	πr2	πr2
Area(segment)	θ    x πr2 360	1   r2  2

S2T3 Solving Simultaneous Equations

Usually given 2 equations with 2 variables, and to find the values of the variables.

Example

Solve for x and y for the following equations:

      3x + 2y = 21  ------ (1) Equation 1

      4x + y = 8 ---------- (2) Equation 2

The variables are x and y

There are usually three methods to use for Simultaneous Equation:

(1) By Substitution

(2) By Elimination

(3) Graphical

(1) By substitution (when there is a X or Y):

Example

Solve for x and y:      

      3x + 2y = 21

        4x + y = 8

Step 1: look for variable with unit digit (1), eg: x or y

Step 2: Arrange equation as such: 

      4x + y = 8

      y = 8 – 4x

Step 3: Substitute y in the other equation (1) and solve

      3x + 2y = 21

      3x + 2(8 – 4x) = 21

      3x + 16 – 8x = 21

        -5x = 21 – 16

       -5x = 5

           x = -1

Substitute x = -1 into y = 8 – 4x, 

          y = 8 -4(-1)

             = 8 + 4 = 12

          y = 12

[[ Test if values of x and y are correct   ]]

Since we use equation 1 to find y, substitute value of x and y into (2).

3 (-1) + 2(12) = -3 + 24 = 21 => value of x and y are correct!

(2) By Elimination

-  'Getting ride' of 1 variable by 'ADDING' or 'SUBTRACTING' both equations

-   1 of the variables must have the same numerical in both equations.

Example

By ADDING

     5y + 2x = 2  --------- (1)

    -5y - 3x = 8  ---------  (2)

  

Using (1) + (2) 

    5y + 2x = 2

+ -5y - 4x = 8.    [5y + (-5y) = 5y - 5y = 0 ]

     0  -x2 = -6  (eliminated Y , find both x and then y)

          x = -6/-2 = 3

Substitute x = 3 into (1) [to find y]

                5y +2(3) = 2

                  y = 2 - 6 = 4

                  y = 4/5

BY SUBTRACTING

       -3x + 2y = 1  ------- (1)

       -3x + 5x = 7 -------- (2)

Using (1) - (2)

       -3x + 2y = 1  

   -   -3x + 5y = 7      [-3x - (-3x) = -3x + 3x]

               -3y = -6 (eliminated X , find y and then x)

                 y = -6/-3 = 2

    Substitute y = 2 into (1)

             -3x + 2(2) = 1

                 -3x = 1 - 4

                    x = -3/-3

                       = 1

Example

Solve the equations by elimination method.

      3x + 2y = 9 --------- (1)  

      4x + 3y = 8 ----------(2)  

Step1: look for variables with smaller digits, (y:2y and 3y, x: 3x, 4x; choose y)

Step2: Make y in both equations to be the same value => 6y by multiplying

      3x + 2y = 9     x3 

      9x + 6y = 27  -------- (3)

          

      4x + 3y = 8    x2  

      8x + 6y = 16   -------- (4)

Step3: Eliminate by + or –

      [both are 6y, 6y => (3) – (4)  because 6y – 6y = 0] 

(3) – (4)

      9x + 6y – (8x + 6y) = 27 – 16

      9x + 6y – 8x – 6y = 11

                  x = 11

OR place the equation in this way:-

        9x + 6y = 27

  -     8x +6y  = 16

          x + 0  = 11   [ x = 9x – 8x, y is eliminated from 6y – 6y= 0, 11 = 27 – 16]

          x = 11

Sub x = 11 into 1

      3(11) + 2y = 9

      33 + 2y = 9

      2y = 9-33

      2y = 24

        y = 12

By Graphical Method

Value of x and y of 2 equations is at the point of interception of the line of equations on the graph 

2 linear equations:y = 2x – 2

                              y = -2x + 4

From the graph, the intersection coordinate is (3/2, 1) => x = 3/2 and y = 1

S2TN Properties and Solving Inequalities

Solving Inequalities                                                            S2 – N7           (S1 – N7 – O)

Recap: 

Symbols

                        >    greater than

                        <    less than

                        ≥    greater than or equal to

                        ≤    less than or equal to

                        =    equal

                        ≠  not equal to

Tip to remember

                                                   4  >  3

        greater = "open mouth"       >      “point” = less than

  

Similarly,

                                                   3  <  4

                  less than = “point”     <       greater = "open mouth"            

Example

    

            Solve 7 < 4x – 3 < 13

 

Step1: Remove number from the expression with x [ 4x – 3]

          [+ 3 to all the expression so that the inequalities remain the same]

 

            7 + 3 < 4x – 3 + 3 < 13 + 3

                10 < 4x < 16

 

Step2: Remove the numerical coefficient of x

                        10 < 4x < 16

                         4     4       4

2 ½ < x < 4

Multiplication property of Inequalities

     x +ve number

               if a < b and c > 0, then ac < bc    eg: 2 < 3 , x 3 => 6 < 9

               If a > b and c > 0, then ac > bc    eg: 3 > 2,  x 3 => 9 > 6

            

      x  –ve number

            If a < b and c < 0, then ac > bc       eg: 2 < 3 , x -3 => -6 > -9 

  => the sign changed from < to >

          

  If a > b and c < 0, then ac < bc       eg: 3 > 2 , x -3 => -6 < -9

  => the sign changed from > to <

   

             [[ x –ve    : change sign from < to > or > to <       ]]

Example

Find the value of x  

a.           5x ≥ 15                                   

               x ≥ 5            ( Divide both side by +5)                    

b.         -2x > 6

              -x < -3               (Divide both side by 2)

             -1 x -x > -3 x -1  (We need to solve for x, and not -x)

               x < 3                 (multiply by -1 => change sign )

S3TN Solving Fractional Equations and Others

Solving Fractional Quadratic Equations

Cross-Multiplying =>

            a = c     =>      x bd  a  =   c   x bd    =>   ad = bc

            b    d                        b       d

[[ The Equation must be ax²+ bx + c = 0 before factorising ]]

Example 

solve    6   = x + 3

          x + 4

            6      =    x + 3 

         x + 4

 6 = (x + 3)(x + 4)         (Step1:  Cross-Multiply)

   6 = x²+ 7x + 12          (Step2: Expand to form a quadratic equation)

   x²+ 7x + 12 – 6 = 0

   x²+ 7x  + 6 = 0

  (x + 1)(x + 6) = 0         (Step3: Factorise and Solve for x)

   x + 1 = 0 or x + 6 = 0

   x = -1 or x = -6

Example

Solve      1       +    2       = 5

            x -2        x – 3 

              (x – 3)  + 2 x (x – 2)   = 5        (Step1: LCM the denominator)

        (x - 3)(x -2)  (x – 2)(x – 3)

            x – 3  + 2x – 4  =  5                   (Step 2 : Combine to single fraction)

              (x – 2)(x -3)

         x + 2x – 3 – 4 = 5        

           (x – 2)(x -3)

               3x – 7       = 5

         (x – 2)(x – 3)

          3x – 7 = 5(x – 2)(x – 3).              (Step 3 : Cross-Multiply and Expand)

          3x – 7 = 5(x²– 5x + 6)

          3x – 7 = 5x²– 5x + 6

          5x²– 5x + 6 – 3x + 7 = 0. 

          5x²-8x + 13 = 0                           (Step 4: Factorise and Solve for x)                                       

          (5x + 13)(x + 1) = 0                                           5x   \ /    13

     5x + 13 = 0 or x + 1 = 0                                           x    / \     1  

           x = -13/5 or x = -1

Solving Equations by equating Coefficients

Example

Find the value of A, B and C

2x⁴ - 13x³ + 19x² + 5x + 1 = (x - 4)(Ax + 1)(x² + Bx + 1) + C

Tips : 
1. Such questions are usually solved by equating the coefficient values
2. If there are more than 2 brackets, expands in pairs
3. Solve the unit value first, and then the highest powers of X

Step 1 - Expand brackets (if needed)

   2x⁴ - 13x³ + 19x² + 5x + 1 = (x - 4)(Ax + 1)(x² + Bx + 1) + C

   2x⁴ - 13x³ + 19x² + 5x + 1 = (x - 4)(Ax + 1)(x² + Bx + 1) + C [Expands 1 pair]

                    (x - 4) (Ax + 1) = (Ax² + (1 - 4)x - 4) (x² + Bx + 1) + C

  2x⁴ - 13x³ + 19x² + 5x + 1 = (Ax² + (1 - 4A)x - 4)(x² + Bx + 1) + C

Step 2 - extract  and solve from comparing coefficients 

  Extract coefficient  of x4.   [=> x² x x² , x³ x x , x⁴ x unit value ]

      2x⁴ - 13x³ + 19x² + 5x + 1 = (Ax² + (1 - 4A)x - 4)(x² + Bx + 1) + C

                   2x⁴  =  Ax² x x²

    A = 2  [Equating x4 coefficient]

    

  Extract and equating coefficient of unit value

    1 = (-4 x 1) + C

C = 1 + 4

C = 5

Step 3 - Substitute values and find remaining unknown

     2x⁴ - 13x³ + 19x² + 5x + 1 = (2x² -7x - 4)(x² + Bx + 1) + 5

  Extract and equating x coefficients

5x = -7x x 1 - 4Bx

             5 = -7 - 4B

  4B = -12

                B = -3 

A = 2, B = -3, C = 5

S2T2 Solving Linear and Fractional Equations

 Solving Linear Equation                                      

Equation : 

(1) a mathematical expression containing the symbol "=" 

           L.H.S = R.H.S

(2) often contains ALGEBRA => variables, such as x and y (to represent the number that we do not know)

(3) examples: y = 3, y + 3 = 10, y = 2x + 3

(4) solving an equation => finding the value of the variable/s (unknown).

(5) must follow the order of operations - BIDMAS ( Bracket, Indices, Divide, Multiply, Add, Subtract).

Example

 Solve for x                                                                            

               5x + 7 = 2x + 28

Step 1 : "move" the variable to LHS, and numbers to RHS

                 LHS   RHS

              5x + 7 = 2x + 28

  5x -2x + 7 - 7 = 2x + 28 - 2x - 7

              5x -2x + 7 - 7 = 2x + 28 - 2x - 7

[ when "moving", the equation must remain 'balance", thus -2x and -7 are applied to both side]

Step2 : Do x/+- to solve

              5x – 2x = 28 – 7 

                      3x = 21

                        x = 7

["Move" also => "changing" the sign of the number/variable at the other side (LHS/RHS).]

Example

Solve for 2y + 4 = y -6

      2y + 4 -y - 4 = y - 6 - y - 4 (step 1: move y to LHS, nos to RHS)

     2y + 4 -y - 4 = y - 6 - y - 4

         y = -10

Solving Fractional Equations (Linear equation)                                  

Linear fractional Equation can usually be solved by simplifying the equation to having a single denominator.

Example

  Solve y/2 + y/3 = 2

Step 1 : Find the lowest common factor (LCM) for the denominators

LCM of 2 and 3 = 6

Step 2 : Combine into single fraction

             ½y + ⅓y = 2

             3 x y + 2 x y  = 2

             3 x 2   2 x 3  

     3y + 2y = 2

                  6

  6 x 5y = 2 x 6 (multiply 6 => "1/6(LHS) = x 6 (RHS))

                  6

  5y = 12

  y = 12/5 = 2 2/5

Example

   Solve y  +  y – 2 = 3

             3        4

 LCM of 3 and 4 = 12

      4x y  +  3x (y – 2)  = 3       (Step 1 :Find LCM and multiply)

      4x 3      3x    4

       

      4y + 3y – 6 = 3       (Step2 : Combine into single fraction)

             12

       7x – 6 = 3        (Step3: Cross-over the denominator and solve)

          12

      7x – 6 = 3 x 12

      7x = 36 + 6

       x = 42/7 = 6

Example

Solve    3   = 3

          x – 2 

          3 = 3(x – 2)    (Step1: Bring the denominator to the right)

            3 = 3x – 6     (Step2: Expand the bracket)

            3x – 6 = 3    (Step3: Solve for x)

            3x = 3 + 6

              x = 9/3

                 = 3