LINES AND ANGLES (LECTURE 5)

30TH APRIL
LESSON  LINES AND ANGLES

DEAR STUDENTS

WELCOME TO THE CLASS!!

FEW REMINDERS:

The text in Red, is to be written in your register

·        

                                                                                                           

1.     Take your  Mathematics Register

2.      Our handwriting reflects a lot about us. It will be awesome if you use good presentation and cursive hand writing

3.      Make a column on the right hand side, if you need to do any rough work

4.      Leave two lines where you finished yesterday’s work and draw a horizontal line

5.      Write today's date on the line after that.

6.      Pending Queries from yesterday (if any).......

7.      Please write the learning outcomes as mentioned below

Today's learning outcomes are  

I will be able to:

1 RECALL THE PROPERTIES OF PARALLEL LINES AND ANGLES FORMED.

2. APPLY THE PROPERTIES LEARNT IN SOLVING QUESTIONS.

Let's revise all the properties done in the previous classes

let's do solved eg 4
Example 4 : In Fig. 6.24, if PQ || RS, ∠ MXQ = 135° and ∠ MYR = 40°, find ∠ XMY .

After the construction the figure will look like ☟

 Now, AB || PQ and PQ || RS.

so, can we say that AB //RS?
Yes, because  lines parallel to the same line are parallel to one another.

Now, ∠ QXM + ∠  XMB = 180° (AB || PQ, Interior angles on the same side of the transversal XM)

But ∠ QXM = 135°   (given)

So, 135° + ∠ XMB = 180°

Therefore, ∠ XMB = 45° (1)

Now, ∠ BMY = ∠ MYR (AB || RS, Alternate angles)

Therefore, ∠ BMY = 40° (2)

Adding (1) and (2), you get

∠ XMB + ∠ BMY = 45° + 40°

That is, ∠ XMY = 85°

↔⭐⭐Important

Example 5 : If a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel, then prove that the two lines are parallel

Given:  A  transversal AD intersects two lines PQ and RS at points B and C respectively. 

 BE is the bisector of ∠ ABQ and  CG is the bisector of ∠ BCS

 and BE || CG.

 To prove that PQ || RS

Look at the figure☟

We are given, BE is the bisector of ∠ ABQ
so, ∠ ABE = ∠ EBQ (SAY ∠1)

Also, CG is the bisector of ∠ BCS

so, ∠ BCG =∠ GCS (SAY ∠ 2)

, BE || CG and AD is the transversal.

Therefore, ∠ ABE = ∠ BCG     (Corresponding angles axiom)  

so,  ∠1  =  ∠2

 therefore 2 ∠1  = 2 ∠2  ( remember the Euclid's Axiom,DOUBLE OF EQUALS ARE EQUAL))


That is, ∠ ABQ = ∠ BCS  

But, they are the corresponding angles formed by transversal AD with PQ and RS; and are equal.

Therefore, PQ || RS

(Converse of corresponding angles axiom) 
LET'S START WITH EX 6.2

Q1 In Fig. 6.28, find the values of x and y and then show that AB || CD

REMEMBER !!

WHEN WE HAVE TO PROVE LINES PARALLEL 

TWO PROPERTIES ARE VERY USEFUL

LINEAR PAIR AXIOM & VERTICALLY OPPOSITE ANGLES

We are given, two lines AB and CD intersected by a transversal
To prove: AB // CD

IN THE FIG.  x +50°  =180°  (why?)

🔴 check the important point mentioned in the beginning of solution)
            =>x  =180°  - 50°
            =>x = 130°


also, y =130° (why?)
🔴 check the important point mentioned in the beginning of solution)
which shows x =y

but, x and y form alternate interior angles

so, we get  AB // CD


Q2. In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.

DONT FORGET TO MARK !!

∠1 IN THE FIG.

since y :z = 3 :7

so, let y =3a and z = 7a where a is any constant

since AB//CD,CD//EF

so, AB//EF

therefore,  ∠ x = ∠ z (alternate interior angles)

so, ∠ x  = 7a

since AB//CD  

we get  ∠ x + ∠ y = 180°   (why?)

3a +7a = 180°

10a= 180°

 a =18°

so, x  = 7a = 7x18° = 126°

GO THROUGH SOLVED EG 6 ON PAGE 103   AND NOTE DOWN IN YOUR NOTEBOOK

REVISE YOUR CW

WE END OUR CLASS HERE!!

SEE YOU FOR THE NEXT CLASS!!

LINES AND ANGLES (LECTURE 4)

PARALLEL LINES AND A TRANSVERSAL.

Good morning boys........

Guidelines for the class:

1. Note down  the work in your register on a regular basis.

2. Please feel free to ask, if you have any doubt by dropping a message in the comment box.

3. Make a column on the right hand side, if you need to do any rough work

4. Write  today's date.

5. Write the chapter number and name.

6. Text in red has to be noted down as its your class work.

7. Text in green has to be read thoroughly and understood.

 Let us recall that a line which intersects two or more lines at distinct points is called a transversal.

Today's learning outcomes are:

• I will be able to comprehend the relation between the angles formed by two or more parallel lines are intersected by a transversal at distinct points.
• Corresponding angles axiom and its converse.
•   Lines parallel to the same line.

Now you see two parallel lines and a transversal..................   

Observe that four angles are formed at each of the intersecting point. C, D,X and W are called the interior angles, while A,B,Y and Z are called exterior angles.

 TRY THESE

ANSWER?

Problem 2

Answer?

Note: Interior angles on the same side of the transversal are also referred to as consecutive interior angles or allied angles or co-interior angles.

Do it in your register 👇

What is corresponding angles axiom?

If a transversal intersects two parallel lines, then each pair of corresponding angles is equal.

This is to read and understand 👇

prove the Converse of the Corresponding Angles Theorem. We will use the very useful technique of proof by contradiction.

We had earlier said axiomatically, with no proof, that if two lines are parallel, the corresponding angles created by a transversal line are congruent.

Now let’s prove an important converse theorem: that if 2 corresponding angles are congruent, then the lines are parallel.

Do it in your register👇

Prove:

If 2 corresponding angles formed by a transversal line intersecting two other lines are congruent, then the two lines are parallel.

To prove this, we will introduce the technique of “proof by contradiction,” which will be very useful down the road.

When we use a “proof by contradiction” method, we start by assuming that what we are trying to prove is NOT true, and we proceed from there until we reach an impossible conclusion – a contradiction.

This means that our original assumption is false (as it leads to this impossible situation), and thus what we were trying to prove must be true.

Do it in your register👇

Proof: Converse of the Corresponding Angles Axiom

So, let’s say we have two lines L1, and L2 intersected by a transversal line, L3, creating 2 corresponding angles, 1 & 2 which are congruent (∠1 ≅ ∠2, m∠1=∠2).

We want to prove the L1 and L2 are parallel, and we will do so by contradiction. Assume L1 is not parallel to L2. Then, according to the parallel line axiom we started with (“for every straight line and every point not on that line, there is one straight line that passes through that point which is parallel to the first line”), there is a different line than L2 that passes through the intersection point of L2 and L3 (point A in the drawing above), which IS parallel to L1.

Let’s draw that line, and call it P. Let’s also call the angle formed by the traversal line and this new line angle 3, and we see that if we add some other angle, call it angle 4, to it, it will be the same as angle 2.

Now, P||L1 so ∠1 ≅ ∠3, as corresponding angles formed by a transversal of parallel lines, and so m∠1=m∠3, from the definition of congruent angles. But we also know that m∠2=m∠3+m∠4, so now perform the substitution m∠2=m∠1+m∠4, and so m∠2 is larger, and not equal to m∠1!

This contradicts what was given,  that angles 1 and 2 are congruent! This contradiction means our assumption (“L1 is not parallel to L2”) is false, and so L1 must be parallel to L2.

LINES PARALLEL TO THE SAME LINE 

Do it in your register👇

  I know that it's bit heavy for you .... so I am not asking you to do any sum as home work, However, you are requested to sit and practice this class work.