Solving Proportionality Rule

In general solving proportionality rule; categorization of proportionality for geometric ratios is a essentially serious way of say what is fundamentally obvious in the specific state of proportional ratios of numbers.  In accumulation, it may be well-known that, given incommensurable magnitudes p and q, this explanation in effect split the field of rational numbers m/n into two disjoint sets: the set L of those for which holds, or m : n < p : q, and the set U of those for which holds, or m : n > p : q.


Rule for Solving Proportionality Rule:

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio for solving proportionality rule.

Solution:

Given:

In preparation for basic proportionality theorem, a triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively.

Rule for Proportionality

To Prove:

(AD)/(DB) = (AE)/(EC)

Construction:

Join BE and CD and then draw DM _|_ AC and EN _|_AB.

Proof:

Area of DeltaADE = (((1)/(2)) base xx height)

So,                   area (DeltaADE) = 1/2AD xx EN

In addition,     area (DeltaBDE) = 1/2 DB xx EN.

Similarly,         area (DeltaADE) = 1/2 AE xx DM and area (DeltaDEC) = 1/2 EC xx DM.

Therefore,        (area ( Delta ADE))/(area ( Delta BDE)) = ( (1/2) AD xx EN )/( (1/2) DB xx EN ) = (AD)/(DB) rArr (i)

And                 (area ( Delta ADE))/(area ( Delta DEC)) = ( (1/2) AE xx DM )/( (1/2) EC xx DM ) = (AE)/(EC) rArr (ii)

Now the DeltaBDE and DeltaDEC is on the same base DE and between the same parallel lines BC and DE.

So,       area (DeltaBDE) = area (DeltaDEC)         rArr (iii)

Therefore, from (i), (ii) and (iii) we have,

(AD)/(DB) = (AE)/(EC).

Hence, the theorem is proved in preparation for solving proportionality rule.


Example for Solving Proportionality Rule:

In solving proportionality rule, figure, AB is parallel to CD. If OA = 3x – 19, OB = x – 4, OC = x – 3 and OD = 4, determine x.

Example for Proportionality

Solution:

Given:

In a quadrilateral ABCD, AB || CD

OA = 3x – 19, OB = x – 4, OC = x – 3, OD = 4

To find:

x = ?

AB || CD

ABCD is a trapezium [The diagonals of trapezium divide each other proportionally]

(3x - 19)/(x - 3) = (x - 4)/(4)

rArr   (x – 4)(x – 3) = 4(3x – 19)

rArr   x^2 - 3x - 4x + 12 – 12x + 76 = 0

rArr   x^2 - 19x + 88 = 0                rArr   x^2 - 11x – 8x + 88 = 0

rArr   x(x – 11) – 8(x – 11) = 0         rArr   (x – 11)(x – 8) = 0

rArr   x – 11 = 0                                or                     x – 8 = 0

x = 11 units or 8 units.

Equality of Condition

Equality of condition states that the given conditions have two or more options that have an equal out come. .

Step  1:Check the given equation of left hand side is equal to right hand side

Step 2:If the left hand side is equal to right hand side the equality condition satisfies.Otherwise it is not a equality of condition(inequality condition)
Solved Examples on Equality of Condition

Example 1: Prove that the given equation satisfies "equality of condition"

3x+7=19

Solution: The Given equation is 3x+7=19

Step 1: Subtract both sides by 7.

3x+7-7=19-7

3x=12

Step 2: Divide both sides by 3

3x/3=12/3

x=4

Step 3:When x=4 only the equality of condition satisfies

Example 2:Check the given equation satisfies "equality of a condition"

(x+5)-(y+9)=4 at x=12 and y=4

Solution: The Given equation is (x+5)-(y+9)=4

Step 1:  Substitute x=12 and y=4 in the equation.

(12+5)-(4+9)=4

17-13=4

4=4

Step 2:So x=12 and y=4 satisfies the equation,this is a equality of a condition.

Example 3:Prove that the given equation satisfies "equality of condition"

4x+5=21

Solution:The Given equation is 4x+5=21

Step 1:Subtract both sides by 5

4x+5-5=21-5

4x=16

Step 2:Divide both sides by 4

4x/4=16/4

x=4

Step 3:When x=4 only the "equality of a condition" satisfies our equation.

Example 4:Check the given equation satisfies "equality of a condition"

(x+2)+(y+1)=9 at x=5 and y=1.

Solution:The given equation is (x+2)+(y+1)=9

Step 1:Substitute x=5 and y=1 in the equation.

(x+2)+(y+1)=9

(5+2)+(1+1)=9

7+2=9

9=9

Step 2:The x and y values satisfies at x=5 and y=1 ,so this is a equality of condition at x=5 and y=1.
Practice Problems on Equality of Condition

Problem 1:Check the given equation is equality of a condition or not?

6x+9=15 at x=1

Problem 2:Check the given equation satisfies "equality of a condition"

(x+3)-(y+1)=9 at x=7 and y=0

Learn Real Roots

In this article we are going to learn about real roots .Roots are the value of the variable that satiafies the given equation.It is also called as sollutions of the equations.The solution may be positive,negative or imaginary numbers.Whenthe roots are real values then its called as real roots. An equation which contains more than one terms are squared but no higher power in terms, having the syntax, ax2+bx+c where a represents the numerical coefficient of x2, b represents the numerical coefficient of x, and c represents the constant term

Example: 4x2+7x+18
Identify the Real Roots in Quadratic Equation

A quadratic equation is in the form of ax2+bx+c,

First we need to find the discernment d = b2- 4ac

If d > 0, the roots are real roots and unequal

If d = 0, the roots are real and equal

If d< 0, the roots are imaginary.
Example of Real Roots

Below you can see the example of real roots -

Example1: Solve the quadratic equation

x2+21x+20

Solution:

The equation is in the form of  ax2+bx+c

Where a=1 and b= 21 and c= 20

Find the discernment d = b2- 4ac

d = (21)2 - 4* 1* 20

d = 441 -80

d = 361 > 0

If d > 0, the roots are real roots and unequal

To find the roots, the following steps are used

Step 1: Multiply the coefficient of x2 and the constant term,

1*20 = 20 (product term)

Step 2: Find the factors for the product term

20--- > 1*20 = 20 (factors are 1 and 20)

20 --- > 1 + 20 = 21 (21 is equal to the coefficient of x)

Step 3: Separate the coefficient of x

x2 + 21x+20

x2 + x + 20x + 20

Step 4: The common term x for the first two terms and 20 for the next two terms are taking outside

x(x + 1) + 20 (x + 1)

(x + 1) (x + 20)

Set this value equal to zero

(x +1) = 0; x = -1

(x+20) = 0; x =-20

Roots of the quadratic equations x =-1 and x= -20

Example2: Solve the quadratic equation

x2+4x+4

Solution:

The equation is in the form of  ax2+bx+c

Where a=1 and b= 4 and c= 4

Find the discernment d = b2- 4ac

d = (4)2 - 4* 1* 4

d = 16 -16

d = 0

If d = 0, the roots are real roots and equal

To find the roots, the following steps are used

Step 1: Multiply the coefficient of x2 and the constant term,

1*4 = 4 (product term)

Step 2: Find the factors for the product term

4--- > 2 * 2 = 4 (factors are 2 and 2)

4 --- > 2 + 2 = 4 (4 is equal to the coefficient of x)

Step 3: Separate the coefficient of x

x2 + 4x+4

x2 + 2x + 2x + 4

Step 4: The common term x for the first two terms and 2 for the next two terms are taking outside

x(x + 2) + 2 (x + 2)

(x + 2) (x + 2)

Set this value equal to zero

(x + 2) = 0; x = -2

(x + 2) = 0; x =-2

Roots of the quadratic equations x =-2.

Problem Solving Situations

The problem solving situations is defined as the process by which the new circumstances situations are collected and resolved. It begins with an solving of all the aspects of the situations problems are solved and ends when a satisfactory answer has been found. The problem solving situations are also functioned in every one life .And now let us see about the problem solving situations.



Problems Based on Problem Solving Situations in Algebra

Determine the given Factor: 144x2 – 81 in term using problem solving situations

Solution:

Step1: The given factor is 144x2 – 81



Step2 : The problem solving situation here is to make squares on both sides means we get the values of 144 = 12 and 81 = 9.

Like 144x2 =81x2

144 = 122

81 = 92

Step3: Arrange the terms 144x2 – 81 = (12x)2 – (9)2      this produced by the different of squares formula

Like, (a + b) (a - b) = a2 + b2 where a is denoted by 12x and b is denoted by  9

Step 4: Therefore the factor of  144x2 – 81 are (12x + 9) (12x – 9).

Find:  8x - 1 = 33  in term using problem solving situations

Solution:

We know about the problem solving situation is to add one on both side we get

8x - 1 + 1 = 31 + 1

After simplifying, 8x = 32.

The required solution are x = 4

One more Problem in Problem Solving Situations Using Algebra

In term using problem solving situations

3x - y = -16---------------------- (1)

2x+8y =-28---------------------- (2)

The above problem handles the problem solving situation as rearranging the equation one means we can get the value for  y .

Step 1: The equation 1 can be written as,

3x - y = -16

y = 3x + 16--------------------- (3)

Now we can assume the value of y in equation 2 means and we get answer for x .

Step 2: Assume the y values in equation (2)

2x + 8(3x + 16) = -28

Step 3: Determine the above equation:

2x + 24x + 128 = -28

26x = -156

x = -6

Step 4: Substitute x values in 1st equation

3(-6) - y = -16

-18-y=-16

Y = -2

The required answer are x = -6, y = -2

The problems can be in term using problem solving situations .

Trigonometric ratios of compound Angles

Definition :

The algebraic sum of two or more angles is called a ' compound angle ' .

If A , B , C are angles , then A + B , A - B , A + B - C  , A - B + C , A + B + C , etc., are compound angles .

Formulas

cos(A + B)  =  cosA cosB  -  sinA sinB  for all A , B in  R .
cos(A - B)   =  cosA cosB  + sinA sinB  for all A , B  in  R .
sin(A + B)   =  sinA cosB  +  cosA sinB  for all  A , B in  R.
sin(A - B)    =  sinA cosB   -  cosA sinB  for all A , B  in  R.
tan(A + B)   =  (tanA+tanB)/(1-tanAtanB)   if  none of A , B , A + B is an odd multiple of (pi)/2  .

cot(A + B)   =  (cotBcotA-1)/(cotB+cotA)   if  none of A , B , A + B is an odd multiple of   (pi)/2 .
tan(A - B)    =   (tanA-tanB)/(1+tanAtanB)    if  none of A , B , A - B is an odd multiple of (pi)/2.
cot(A - B)    =   (cotBcotA+1)/(cotB-cotA)    if  none of A , B , A + B is an odd multiple of  (pi)/2.
sin(A + B) sin(A - B)   =  sin2A - sin2B    =    cos2B  -  cos2A .
cos(A + B) cos(A - B)   =  cos2A - sin2B   =   cos2B  -  sin2A


Solved Problems on Trigonometric Ratios of Compound Angles

1) Find the value of sin 105o and cos 165o .

Sol: sin 105o   =   sin (45o + 60o)

=    sin45o  cos60o   +  cos45o sin60o

=   (1/sqrt(2)) (1/2)   +  (1/sqrt(2)) (sqrt(3)/2)

sin105o   =   (sqrt(3)+1)/(2sqrt(2))

Now , cos165o   =   cos (180o - 15o)    =   - cos15o

=   - cos(45o - 30o)

=   - [ cos45o cos30o + sin 45o sin 30o]

=   -[(1/sqrt(2))(sqrt(3)/2)+(1/sqrt(2))(1/2)]

cos 165o   =   -(sqrt(3)+1)/(2sqrt(2))

2) Show that   cos42o + cos78o + cos162o = 0 .

Solution :  cos42o + cos78o + cos162o

=   cos (60o - 18o)  +  cos(60o + 18o)  +  cos(180o - 18o)

=   (cos60o cos18o + sin60o sin18o)  +  (cos60ocos18o - sin60o sin18o)  -  cos18o

=   2 cos60ocos18o  -  cos18o

=   2(1/2) cos18o  -  cos18o  =  0


More Problems on Trigonometric Ratios of Compound Angles:

1)Find the values of sin 15o , cos 15o , tan15o

Sol:

sin15o   =  sin (60o - 45o)   =  sin 60o cos45o  -  cos60o sin45o

=  (sqrt(3)/2)(1/sqrt(2)) -  (1/2)(1/sqrt(2))

sin 15o  =  (sqrt(3)-1)/(2sqrt(2))

cos 15o   =   cos(60o - 45o)   =   cos60o cos45o  +  sin60o sin45o

=   (1/2)(1/(sqrt(2)))   +  (sqrt(3)/2)(1/sqrt(2))

cos15o     =   (sqrt(3)+1)/(2sqrt(2))

tan15o   =   tan(60o - 45o)   =   (tan60^o-tan45^o)/(1+tan60^otan45^o)

=    (sqrt(3)-1)/(2sqrt(2))

tan15o   =   2 -  sqrt(3)

2) Show that  cos100o cos40o  +  sin100o sin40o   =  1/2

Solution :  cos100o cos40o  +  sin100o sin40o

The above equation is in the form of  cosA cosB  +  sinA sinB  which is equal to cos(A - B)

Here A = 100o  and  B = 40o

cos(A - B)  =  cos(100o - 40o)

=  cos(60o)

=  1/2

Hence  cos100o cos40o  +  sin100o sin40o  =  1/2

Solve Two Pairs of Angles Problems

Let us see about solve two pairs of angles problems in this article. Two lines are intersecting to make a two pair of angle. Two lines sharing the common end point is representing as angle. The word angle is derived form the Latin word. In Geometry angle is the one of the figure. In two pair of angle opposite angle are same and parallel.
Two Pairs of Angle:

There are several types of two pairs of angles is available for the geometry. These are

Complementary angle
Supplementary angle
Vertical angle

Example Diagram for Solve Two Pairs of the Angles Problems:

Pair of complementary angle:

The sum of the two angle measurement is equal to 90 is called as complementary angle.

example figure for two pair of the angles

Pair of supplementary angle:

The sum of the two angle measurement is equal to 180 is calling as complementary angle.

example figure for supplementary pair of angles

Pair of vertical angle:

One line crossed by another line is formed by the intersecting is said to be vertical angle.

example figure for vertical pair of angles
Example 1: Solve Two Pairs of Angles Problems:

In the given diagram find the unknown value of the a, b, c.

example for two pair of angle diagram

Solution:

Step 1: angle x is supplement of 80°

Step 2: a + 80° = 180°

a = 180° - 80°

a = 100°

Step 3: In this diagram angle c and 100° are vertical angles.

So, angle c = 100°

Step 4; In this diagram angle b and 80° are vertical angle

So, angle b = 80°

Step 5:

a = 100°

b = 80°

c = 100°


Example 2: Solve Two Pairs of Angles Problems:

In complementary angle, angle A is 37 degree calculate the unknown angle

Solution:

Step 1: Total angle measurement of the complementary angle is 37°

Step 2: Given angle is A = 37°

Step 3: Subtracting given angle measurement from total angle measurement of complementary angle. 90° - 37° = 53°

Step 4: Unknown angle is 53°

Problem Solving Training

In this article we are giving problem solving training and we can understand how to solving problems. It is very helping you to improve your problem solving skills. In mathematical terms we can see different types of problem solving. Here you can learn math terms, Exam practices test, problem solving and online test and our tutor will helps you and you can get free online tutor. Let us see Problem solving training.
Problem Solving:

Let us see few problems and their solving methods.

Problem solving training 1:

Solve:  `12 / 6 + 5 / 3`

Solution:

Above problem is showing fraction addition. So we should solve this problem in fraction addition operation.

Step 1: `12 /6 + 5 / 3`

Here numerator values are same, but denominators are different. So we should take LCM then only we can add both values. (We can take LCM if denominators are different).

Step 2:  `12 / 6 + 5 / 3`

Take LCM 6, 3 (therefore LCM is 6)

we have to change denominators values like 6.

Step 3: `(12*1)/(6*1) = 12/6 , (5*2)/ (3*2) = 10 / 6`

`12/6 + 10 / 6`

Now denominators are same so add both values

Step 4: `12 / 6 + 10 / 6`

`22 / 6`

Therefore `12/ 6 + 5 / 3 = 22 / 6.`

Problem solving training 2:

Solve:    20 ___ 5   = 25

Solution:

Step 1: Given 20 ___ 5   = 25.

Step2: here we find the symbols which are need for this operation.

Step 3: if we put the - (minus) symbols like 20 - 5 = 15 we can get 15.

So minus operation is not accept

Step4: + (plus) is correct operation for this problem.

20 + 5 = 25

Step 5: Therefore + (plus) symbol is making the number sentences true.

Problem solving training 3:

Divide the two fractions `40 -: 1/ 5`

Solution:

Above problem is showing fraction division. So we should solve this problem in fraction division operation.

Step 1: given` 40 -: 1/ 5`

Step 2: It denoted by `40 -: 1/5`

The right hand side denominator will be change like as 5/1 so

=   `40 * 5/1`

= `200`

Step 3: Therefore answer is 200.
Practices Problems:

1) Solve this fraction `40/ 20`      answer: 2

2) Add `2 / 3 + 3/ 2 `                      answer: `13/6`

3) Find missing number 4 , 8  12 , ___ , 20 , 24 , ____ , 32      answer: 16 , 28