**Lines and Angles:**

Suppose we have a ray OA on a plane with end point O. With the same plane end point. Consider another ray OB in the same plane. We observe that OB is obtained from OA through suitable rotation around the point O. We say OB subtends an angle with angle OA. The amount of rotation is the **measure **of this angle.

∠AOB

We use numerical measurement called **degree** to measure angles. We use the notation **a˚** to denote **a degrees. **The rays OA and OB are called the **sides **of the angle and O is called the **vertex** if the angle. The angle subtended by the rays OA and OB is denoted by ∠AOB.

We know a geometrical instrument called **protractor **which is used to measure the angles.

We Know, different types of angles:

**Straight line angle, right angle, acute angle, obtuse angle, reflex angle, complete angle, adjacent angle, complementary angles and supplementary angles.**

Let us recall one by one,

Two angles are said to be **supplementary angles **if their sum is 180˚. Similarly, two angles are said to be **complementary** if they add up to 90˚

Two angles are said to be **adjacent angles**, if both the angles have a common vertex and a common side.

While measuring the lengths are segments of the angles we observe the following rules:

Rule 1:Every line segment has a positive length.(The length of the line segment AB is denoted by AB or |AB|.)

Rule 2:If a point C lies on a line segment |AB|, then the length of |AB| is equal to the sum of the lengths of |AC| and |CB|; that is AB = AC + CB.

Rule 3:Every angles has a certain magnitude. A straight angles measures 180°

Rule 4:If OA, OB and OC are such that OC lies between OA and OB then ∠AOB = ∠AOC + ∠COB

Rule 5:If the angle between two rays is zero then they coincide. Conversely, if two rays coincide, the angle between them is either zero or an integral multiple of 360°

__Theorem 6.1 :__**If two lines intersect each other, then the vertically opposite angles are equal.**

__Proof :__

In the statement above, it is given that ‘two lines intersect each other’. So, let AB and CD be two lines intersecting at O as shown in Fig. 6.8. They lead to two pairs of vertically opposite angles, namely,

(i) ∠ AOC and ∠ BOD (ii) ∠ AOD and ∠ BOC.

We need to prove that ∠ AOC = ∠ BOD and ∠ AOD = ∠ BOC.

Now, ray OA stands on line CD.

Therefore, ∠ AOC + ∠ AOD = 180° *(Linear **pair axiom)* ………..(1)

Can we write ∠ AOD + ∠ BOD = 180°? *(Linear **pair axiom)*……………(2)

From (1) and (2), we can write

∠ AOC + ∠ AOD = ∠ AOD + ∠ BOD

This implies that ∠ AOC = ∠ BOD

Similarly, it can be proved that ∠AOD = ∠BOC

** Example 1 :** In Fig., lines PQ and RS intersect each other at point O. If ∠ POR : ∠ ROQ = 5 : 7, find all the angles.

** Solution :** ∠ POR +∠ ROQ = 180°

*(Linear pair of angles)*

But ∠ POR : ∠ ROQ = 5 : 7 *(Given)*

Therefore, ∠ POR = ^{5}/_{12} × 180° = 75°

Similarly, ∠ ROQ = ^{7}/_{12} × 180° = 105°

Now, ∠ POS = ∠ROQ = 105° *(Vertically opposite angles)*

*and *

∠ SOQ = ∠POR = 75° *(Vertically opposite angles)*

__Example 2 :__

In Fig., ray OS stands on a line POQ. Ray OR and ray OT are angle bisectors of ∠ POS and ∠ SOQ, respectively. If ∠ POS = x, find ∠ ROT.

** Solution :** Ray OS stands on the line POQ.

Therefore, ∠ POS + ∠ SOQ = 180°

But, ∠ POS = x

Therefore, x + ∠ SOQ = 180°

So, ∠ SOQ = 180° – x

Now, ray OR bisects ∠ POS, therefore,

∠ ROS = ^{1}/_{2} × ∠ POS = ^{1}/_{2} × x = ^{x}/_{2}

Similarly, ∠ SOT = ^{1}/_{2} × ∠ SOQ = ^{1}/_{2} × (180° – x)

= 90˚ – ^{X}/_{2}

Now, ∠ ROT = ∠ ROS + ∠ SOT

= ^{x}/_{2} + 90° – ^{x}/_{2}

= 90°

__Example 3 :__

In Fig. 6.11, OP, OQ, OR and OS are four rays. Prove that ∠ POQ + ∠ QOR + ∠ SOR + ∠ POS = 360°.

** Solution :** In Fig. 6.11, you need to produce any of the rays OP, OQ, OR or OS backwards to a point.

Let us produce ray OQ backwards to a point T so that TOQ is a line (see Fig. 6.12).

Now, ray OP stands on line TOQ.

Therefore, ∠ TOP + ∠ POQ = 180° …………..(1) *(Linear pair axiom)*

Similarly, ray OS stands on line TOQ.

Therefore, ∠ TOS + ∠ SOQ = 180° ……………(2)

But ∠ SOQ = ∠ SOR + ∠ QOR

So, (2) becomes

∠ TOS + ∠ SOR + ∠ QOR = 180° ………………….(3)

Now, adding (1) and (3), you get

∠ TOP + ∠ POQ + ∠ TOS + ∠ SOR + ∠ QOR = 360°…………………(4)

But ∠ TOP + ∠ TOS = ∠ POS

Therefore, (4) becomes

∠ POQ + ∠ QOR + ∠ SOR + ∠ POS = 360°

**EXERCISE 6.1**

- In Fig., lines AB and CD intersect at O. If ∠ AOC + ∠ BOE = 70° and ∠ BOD = 40°, find ∠ BOE and reflex ∠ COE.

- In Fig., lines XY and MN intersect at O. If ∠ POY = 90° and a : b = 2 : 3, find c.

- In Fig., ∠ PQR = ∠ PRQ, then prove that ∠ PQS = ∠ PRT.

- In Fig., if x + y = w + z, then prove that AOB is a line.

- In Fig., POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ ROS =
^{1}/_{2}(∠ QOS – ∠ POS).

- It is given that ∠ XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ ZYP, find ∠ XYQ and reflex ∠ QYP.

**Lines and Angle – Exercise 6.1**

**Parallel Lines and a Transversal**

Recall that a line which intersects two or more lines at distinct points is called a transversal. Line l intersects lines m and n at points P and Q respectively. Therefore, line l is a transversal for lines m and n. Observe that four angles are formed at each of the points P and Q.

Let us name these angles as ∠ 1, ∠ 2, . . ., ∠8 as shown in Fig. ∠ 1, ∠ 2, ∠ 7 and ∠ 8 are called exterior angles, while ∠ 3, ∠ 4, ∠ 5 and ∠ 6 are called **interior angles**.

Recall that in the earlier classes, you have named some pairs of angles formed when a transversal intersects two lines. These are as follows:

(a) Corresponding angles:

(i) ∠ 1 and ∠ 5

(ii) ∠ 2 and ∠ 6

(iii) ∠ 4 and ∠ 8

(iv) ∠ 3 and ∠ 7

(b) Alternate interior angles :

(i) ∠ 4 and ∠ 6

(ii) ∠ 3 and ∠ 5

(c) Alternate exterior angles:

(i) ∠ 1 and ∠ 7

(ii) ∠ 2 and ∠ 8

(d) Interior angles on the same side of the transversal:

(i) ∠ 4 and ∠ 5

(ii) ∠ 3 and ∠ 6

Interior angles on the same side of the transversal are also referred to as consecutive interior angles or allied angles or co-interior angles. Further, many a times, we simply use the words alternate angles for alternate interior angles.

**Exercise 6.2 :**

- In Fig. 6.28, find the values of x and y and then show that AB || CD.
- In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.
- In Fig. 6.30, if AB || CD, EF ⊥ CD and ∠ GED = 126°, find ∠ AGE, ∠ GEF and ∠ FGE.
- In Fig. 6.31, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠ QRS. [Hint : Draw a line parallel to ST through point R.]
- In Fig. 6.32, if AB || CD, ∠ APQ = 50° and ∠ PRD = 127°, find x and y.
- In Fig. 6.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.

**Lines and angles – Exercise 6.2**

**Lines and Angles – EXERCISE 6.3 **

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