Q1. In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

 

 

 

 

 

Answer:

We are given

∠AOC + ∠BOE = 70°

∠BOD = 40°

∠BOD = ∠AOC (Vertically opposite angle)

∴ ∠AOC = 40°

Putting values of  ∠AOC in the given equation ∠AOC + ∠BOE = 70°

40 + ∠BOE = 70

∠BOE = 30°

∠AOE + ∠BOE = 180° (Linear pair)

∠AOC + ∠COE + ∠BOE = 180°

40° + ∠COE + 30° = 180°

∠COE = 180° – 70° = 110°

Reflex ∠COE = 360° – ∠COE

Reflex ∠COE = 360° – 110° =250°

Therefore  ∠BOE =30° and reflex ∠COE =250°

Q2. In Fig. 6.14, lines XY  and MN intersect at O. If ∠POY =90° and a:b=2:3, find c.

 

 

 

 

 

 

Answer:

We are given

∠POY =90° and a: b =2: 3

∠POX = 180° – 90° = 90°(Linear pair)

Let a = 2x and b = 3x

∠POX = a + b = 90

2x + 3x = 90

5x = 90°

x = 18°

Therefore a = 2×18 = 36° and b = 3× 18 = 54°

Since MON is a line therefore b and c are linear pair

∴b + c =180°

54° + c = 180°

c = 180° – 54°  = 126°

Hence the value of c is 126°

Q3. In fig. 6.15 ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT

 

 

 

 

 

Answer:

Given: ∠PQR = ∠PRQ

To Prove: ∠PQS = ∠PRT

Proof: In the given fig.

∠PQS + ∠PQR = 180° (Linear pair)…..(i)

∠PRT + ∠PRQ = 180° (Linear pair)…..(ii)

From equations (i) and (ii)

∠PQS + ∠PQR = ∠PRT + ∠PRQ

∠PQR = ∠PRQ (given)

∠PQS + ∠PRQ = ∠PRT + ∠PRQ

∠PQS = ∠PRT

Hence Proved

Q4. In fig.6.16 if x +y = w +z, then prove that AOB is a line.

 

 

 

 

 

 

 

 

Answer: We have In the given figure

Given: x +y = w +z

To Prove: AOB is a line

Proof: From the figure, we can observe that all the given angles form a complete angle of 360°

Therefore we can write

x +y + w +z = 360° (complete angle)

w + z + w + z = 360°

2w + 2z = 360°

2(w + z) = 360°

w + z = 180

The sum of w and z implies that both angle w and z forms a linear pair, so AOB is a line.

Q5. 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 = 1/2(∠QOS -∠POS).

 

 

 

 

 

Answer:

Given: POQ is a line

OR ⊥ PQ

∴ ∠POR = 90° and ∠QOR = 90°

To Prove: ∠ROS = 1/2(∠QOS -∠POS)

Proof: POQ is a line (given)

Therefore ∠POR  and ∠QOR are the linear pair

∠POR  +  ∠QOR = 180°

∠POS + ∠ROS + 90° = 180°

∠POS + ∠ROS = 90°…..(i)

From the fig. we have

∠QOS  = ∠QOR + ∠ROS

∠QOS  = 90° + ∠ROS

∠QOS – ∠ROS = 90°……(ii)

From (i) and (ii)

∠POS + ∠ROS = ∠QOS – ∠ROS

∠ROS + ∠ROS  = ∠QOS – ∠POS

2∠ROS   = ∠QOS – ∠POS

∠ROS = 1/2(∠QOS – ∠POS)

Hence proved

Q6. 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.

Answer:

 

 

 

 

 

We are given

∠XYZ = 64°

Let ∠PYZ = 2x

∠QYZ = ∠QYP = x ( YQ is the bisector of ∠PYZ)

Since PYX is a line,so

∠PYZ +∠XYZ = 180°

∠QYZ + ∠QYP + 64° = 180°

x + x = 180° – 64° = 116°

2x = 116°

x = 58°,so ∠QYZ = ∠QYP = 58°

∴∠XYQ = ∠XYZ + ∠QYZ  = 64° + 58° = 122°

Since  ∠QYP = 58°, therefore reflex ∠QYP = 360°-58° = 302°

Hence ∠XYQ = 122° and reflex ∠QYP  = 302°