S2 - G4 : Pythagoras' theorem and Trigonometry

Pythagoras' Theorem

For a right angle triangle:-

If the hypotenuse has length c, and the sides have lengths a and b, then

c² = a² + b²  

[hypotenuse = the length across or opposite the right angle / longest length]

=> if the sides of a triangle have lengths a, b and c, such that c² = a² + b², 

      then the triangle is a right-angle triangle.

Example 

What is c?

Using c² = a² + b²

                  5² + 12²

                =  25 + 144

                = √169

                = 13

Example         

If every square is 1 unit, what is a?

 Using c² = a² + b² 

           25 = 5², 9 = 3²  

           5² = a² + 3²   

           a²  = 25 - 9

           a²  = √16
             a = 4

Trigonometric Ratios

  Tangent ,Cosine and Sine

    To find angle x or any of the sides, we can use   

               Tan x  = Opposite   =  b                 TOA

                             Adjacent       a

                Cos x =   Adjacent  = a                 CAH     

                             Hypotenuse   c

               

                Sin x =    Opposite   = b                SOH

                            Hypotenuse    c

Example

Find 

     (a) length AC

     (b) Sin X

     (c) Cos X

     (d) Tan X

     (e) Angle X

a. Length AC

      c² = a² + b²            Step 1 : which formula to use? Label a=6, b = 8

                                                  [Right angle ◺, use Pythagoras Theorem]

      c² = 6² + 8²            Step 2  : Compute and answer

      c = √36 + 64  

         = 10

    AC = 10 cm

b. Sin X =  Opp/Hyp     Step 1 : Determine the formula and label 

             = 6/10              Step 2 : Compute and answer  

             = 3/5              

c.  Cos X = Adj/Hyp      Step 1 : Determine the formula and label

             = 8/10              Step 2  : Compute and answer

             = 4/5               

d.  Tan X = Opp/Adj

               = 6/8 = 3/4

e. Tqn X = 6/8               Step 1 : Determine the formula and label

          X = Tan^-1(6/8)      Step 2  : Compute and answer

             =  36.86^o

[If possible, always use the value given to do the computation instead of a computed value. This is to avoid wrong answer in case there is of a mistake in the computed value]

Angle of Elevation and Depression

The angle is measured between the horizontal line (AB) and the line forming the angle. 

    

            Angle of Elevation – Looking Up

            Angle of Depression – Looking Down

Sine Rule

            a       =     b     =    c

          SinA        SinB      SinC                 

                         or 

          SinA  =  SinB   =   SinC

             a            b             c

Cosine Rule

             a² = b² + c² – 2bcCosA

Note:

1. The value of Sinθ and Cosθ is between -1 to 1 =>

                    -1  ≤ Sinθ ≤ 1

                    -1  ≤ Sinθ ≤ 1

    Check your computation if your answer for Sinθ orCosθ is not within -1 to 1

2.  For Pythagoras Theorem, the Hypotenuse length is the longest 

     => the value of both sides are smaller than the hypotenuse length.

     Check your computation  if any side has a bigger value than the hypotenuse. 

3.  If possible, always use the given values in the question for computation.

Practice

Find 

     (a) length BC

     (b) Sin Y

     (c) Cos Y

     (d) Tan Y

     (e) Angle Y

     (f)  Sin Z

     (g) Cos Z

     (h) Tan Z

     (i)  Angle Z

Answer

a. AC² = AB² + BC²        Step 1 : Right Angle ◺, use Pythagoras T]

      13² = 5² + BC²         Step 2 : Label Diagram and Compute

      BC² = 13² - 5²

      BC = √169 - 25

            = √144 

            = 12

b.  Sin Y = O/H               Step 1 : Determine the formula and labe

               = 12 / 13          Step 2 : Label Diagram and Compute

c.  Cos Y = A/H

               = 5/13

d.  Tan Y = O/A

               = 12/5

e.   Y = Cos ^-1 (5/13)

         = 67.38^o   

f.   Sin Z  = 12/13

g.  Cos Z = 5/13

h.  Tan Z = 5/12

i. Angle Z = 180 - 90 - 67.38^o     

                = 22.62^o   

[If possible, always use the given values in the question; this is to avoid subsequent wrong answers if there is error in a computed value.]

N63-G23 Quadratic Functions and Graphs

Graph of Quadratic Functions

(1)  Of the form y= ax² + bx + c ( b and c can be zero, but a ≠ 0)

(2) All have a line of symmetry

(3) U-shaped

(4) Turning Point:

      (i) Positive [+ax²] graph - turning point at the bottom (Minimum point)

     (ii) Negative [-ax²] graph - turning points at the top (maximum) 

Plotting the Graph

Example (1) 

(a) Complete the table for y = -x² + x -3 for -2 <= x <=3.

X	-2	-1	0	1	2	3
Y	-9

(b) Plot and draw the graph.

(c) Use the graph to solve the equations 

     (a) –x² + x – 3 = -7

     (b)  –x² + x – 3 = 0

(a) Filling in the table

(a) The table is for the x and y values of equation y = –x² + x – 3

Step 1:  Write the equation, and substitute x value into y = –x² + x – 3.

Starting with x = -2

                        y = –x² + x – 3

                        y = -(-2)² + (-2) – 3

                        y = -(4) – 2 – 3

                           = -4 -2 – 3

                           = -9

Do for other x values:-

            When x = - 1, y = -(-1)² + (-1) – 3 = -5

            When x = 0, y = -(0)² + (0) – 3 = -3

            When x = 1, y = -(1)² + (1) – 3 = -3

            When x = 2, y = -(2)² + (2) – 3 = -5

            When x = 3, y = -(3)² + (3) – 3 = -9

Step 2: Fill in the values 

X	-2	-1	0	1	2	3
Y	-9	-5	-3	-3	-5	-9

(b) Drawing the Graph

(b) Plot and draw the graph.

The graph is plotted by placing and joining the coordinates on the graph.

(c) Solving Equation using the Graph

There are two variables ( x and y) in the equation y = ax² + bx + c. 

<<a, b, c are constant (a specific number) >>

We can solve an equation when either x or y is given. 

When y is given => find x, 

When x is given => find y

When using a graph to solve an equation ax² + bx + c, 

Step 1 : Form an equation such that

    y = ax² + bx + c = k

=> y = k

Step 2 : Draw y = k on the graph

Step 3 : Solve the equation by locating the intersection of y = k and y = ax² + bx + c and find the required value/s. 

(c) Use the graph to solve the equation 

      (i)  –x² + x – 3 = -7

Step 1: Write and form the equation

 (i)         y =  –x² + x – 3 = -7

Step 2: ‘Equate’ both to get equation to plot/draw on graph

           y = -7

Step 3: Draw the line and find the coordinate               

                  Draw y = -7

From the graph, x = -1 2/5 and 2 2/5

(ii) –x² + x – 3 = 0 

        y =  –x² + x – 3 = 0 (Step 1) 

        Y = 0                       (Step 2)

       Draw y = 0               (Step 3)

Since y = 0 and y =  –x² + x – 3 do not "cut" or intersect, the values are undefined => (no intersections of the line and equation).

Example(2)

Use the graph of y = 3x² – 2x + 4 to find

      (a) 3x² – 2x + 4 = 12

                 y =  3x² – 2x + 4 = 12  (Step 1)

                 y = 12                          (Step 2)

Draw the line and find coordinate(Step 3)

          x = -4/3 or x = 2  

(b) 3x² – 2x – 3  = 0

Step 1: Write the equation

      y =  3x² – 2x + 4

      y =  3x² – 2x - 3 = 0  (Step 1)

Step 2: Equate 

[need to ‘change’ equation to 3x² – 2x + 4]

To equate, +7 needs to be "added" 

       y = 3x² – 2x - 3 +7 = 0 + 7

       y= 3x² – 2x + 4 = 7

           y = 7

Step 3: Draw the line y = 7 and find coordinate

            x = -0.72 or x = 1.4

Finding Minimum/Maximum Points

Example:  Find the maximum and minimum value of y = -x² + 6x + 6 and y = x² + 4x + 5

Step 1: Find the value of x when y = 0

  y = -x² + 6x + 6 y = x² + 4x - 5

     When y = 0 Step 1: find the value of x when y = 0

-x² + 6x + 6 =0 x² + 4x - 5 = 0

     (x + 1)(x -7) = 0 (x + 5)(x - 1) = 0  (by factorisation)

      x = -1, or x = 7 x = -5, x = 1

Step 2: Find the mid-point for the values of x

Mid point of x = -1 and x = 7 Mid point of x = -5 and x = 1

   |--|--|--|--|--|--|--|--|-   -|---|---|---|---|---|---|--

  -1 0 1 2  3 4 5 6 7 -5 -4 -3  -2 -1  0   1

From the number line = 3 From the number line = -2

Step 3: Find the value of y

Value of y: 

y = -(3)² +6(3) + 6  y = (-2)² + 4(-2) - 5

  = -9 + 18 + 6     = 4 - 8 - 5

              = 15     = -9

 

Step 4: List the maximum/minimum point

     Mid-point = (3, 15)   Mid-point = (-3, -9)

By graphical method:

Example:  Find the maximum and minimum value of y = -x² + 6x + 6 and y = x² + 4x + 5

Summary

Practice

Using Example(2)

(a)        Find the values of x for 3x² – 2x + 4 = x + 6 

(b)        Find the values of x for 3x² – 3x – 2  = 0