**Exercise: 6.1 (Page No: 96)**

**1. In Fig. 6.13, lines AB and CD intersect at O. If AOC +BOE = 70° and BOD = 40°, find BOE and reflex COE.**

**Solution:**

From the diagram, we have

(AOC +BOE +COE) and (COE +BOD +BOE) forms a straight line.

So, AOC+BOE +COE = COE +BOD+BOE = 180°

Now, by putting the values of AOC+BOE = 70° and BOD = 40° we get

COE = 110° and

BOE = 30°

**2. In Fig. 6.14, lines XY and MN intersect at O. If POY = 90° and a : b = 2 : 3, find c.**

**Solution:**

We know that the sum of linear pair are always equal to 180°

So,

POY +a +b = 180°

Putting the value of POY = 90° (as given in the question) we get,

a+b = 90°

Now, it is given that a : b = 2 : 3 so,

Let a be 2x and b be 3x

∴ 2x+3x = 90°

Solving this we get

5x = 90°

So, x = 18°

∴ a = 2×18° = 36°

Similarly, b can be calculated and the value will be

b = 3×18° = 54°

From the diagram, b+c also forms a straight angle so,

b+c = 180°

c+54° = 180°

∴ c = 126°

**3. In Fig. 6.15, PQR = PRQ, then prove that PQS = PRT.**

**Solution:**

Since ST is a straight line so,

**∠**PQS+**∠**PQR = 180° (linear pair) and

**∠**PRT+**∠**PRQ = 180° (linear pair)

Now, **∠**PQS + **∠**PQR = **∠**PRT+**∠**PRQ = 180°

Since **∠**PQR =**∠**PRQ (as given in the question)**∠**PQS = **∠**PRT. (Hence proved)

**4. In Fig. 6.16, if x+y = w+z, then prove that AOB is a line.**

**Solution:**

For proving AOB is a straight line, we will have to prove x+y is a linear pair

i.e. x+y = 180°

We know that the angles around a point are 360° so,

x+y+w+z = 360°

In the question, it is given that,

x+y = w+z

So, (x+y)+(x+y) = 360°

2(x+y) = 360°

∴ (x+y) = 180° (Hence proved).

**5. In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ROS = ½ (QOS – POS).**

**Solution:**

In the question, it is given that (OR ⊥ PQ) and POQ = 180°

So, POS+ROS+ROQ = 180°

Now, POS+ROS = 180°- 90° (Since POR = ROQ = 90°)

∴ POS + ROS = 90°

Now, QOS = ROQ+ROS

It is given that ROQ = 90°,

∴ QOS = 90° +ROS

Or, QOS – ROS = 90°

As POS + ROS = 90° and QOS – ROS = 90°, we get

POS + ROS = QOS – ROS

2 ROS + POS = QOS

Or, ROS = ½ (QOS – POS) (Hence proved).

**6. 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.**

**Solution:**

Here, XP is a straight line

So, XYZ +ZYP = 180°

Putting the value of XYZ = 64° we get,

64° +ZYP = 180°

∴ ZYP = 116°

From the diagram, we also know that ZYP = ZYQ + QYP

Now, as YQ bisects ZYP,

ZYQ = QYP

Or, ZYP = 2ZYQ

∴ ZYQ = QYP = 58°

Again, XYQ = XYZ + ZYQ

By putting the value of XYZ = 64° and ZYQ = 58° we get.

XYQ = 64°+58°

Or, XYQ = 122°

Now, reflex QYP = 180°+XYQ

We computed that the value of XYQ = 122°.

So,

QYP = 180°+122°

∴ QYP = 302°