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Exercise 6.1

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

Solution:

 ∠BOD = 40°

∠AOC  = ∠BOD (Vertically opposite Angle)

∠AOC = 40°

∠AOC  + ∠ BOE = 70° (Given)

∠BOE = 70°

∠BOE = 70° - 40°

∠BOE = 30°

AOB is straight line

∠AOC +  ∠COE +∠BOE = 180° (linear pair)

⇒ 70° + ∠COE = 180°

⇒ ∠COE = 180° - 70°

⇒ ∠COE = 110°

Reflex ∠COE = 360 - 110°

                      = 250°

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

Solution:

∠POY=90° (given) 

Let  ∠a and ∠b = 2x and 3x

XOY is a straight line

∠a + ∠b + ∠POY = 180°

⇒2x + 3x + 90°= 180°

⇒5x  = 180° ­­- 90°

⇒5x = 90°

⇒x = 90°/5

⇒x = 18°

Now ∠a = 2 x 18°

             = 36°

         ∠b =3 x 18°

               = 54°

MON is a straight line   

∠b + ∠c = 180°(linear pair)

∠54° + ∠c = 180°

⇒∠c = 180°- 54°

          =126°

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

Solution : 

Given : ∠PQR = ∠PRQ

To prove : ∠PQS = ∠PRT

Proof : 

∠PQS + ∠PQR = 180°  .................. (1)  Linear pair

∠PRT + ∠PRQ = 180°  .................. (2)  Linear pair

From equation (1) and (2) 

       ∠PQS + ∠PQR = ∠PRT + ∠PRQ 

Or,  ∠PQS + ∠PQR = ∠PRT + ∠PQR    (∠PQR = ∠PRQ given) 

Or, ∠PQS = ∠PRT Proved 

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

Solution: 

Given : x + y = w + z 

To prove : AOB is a line.

Proof :  

We know that;

x + y + w + z = 360^०

(Angle Sustained on centre) 

x + y + x + y =  360^० (x + y = w + z given) 

2x + 2y = 360^० 

2 (x + y) = 360^० 

x + y = 180^० (linear pair) 

Therefore, AOB is a line   

Hence, Proved

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

Solution:

Given:

POQ is a straight line. OR ⊥ PQ and OS is

another ray lying between rays OP and OR.

To prove: 

Proof: OR ⊥ PQ  (given)

∴ ∠QOR = 90^०  …………… (1)

 POQ is straight line

∴ ∠POR + ∠QOR = 180^० (linear pair)

⇒ ∠POR + 90^० = 180^०

⇒ ∠POR = 180^०– 90^०

⇒ ∠POR = 90^०…………… (2)

Now,  ∠ROS + ∠QOR = ∠QOS

Or,       ∠ROS = ∠QOS – ∠QOR  ……………. (3)

Again, ∠ROS + ∠POS = ∠POR

Or,       ∠ROS = ∠POR – ∠POS  ……………. (4)

Adding equation (1) and (2)

∠ROS + ∠ROS = ∠QOS – ∠QOR + ∠POR – ∠POS 

2 ∠ROS = ∠QOS – 90^०+ 90^०– ∠POS

 2 ∠ROS = (∠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.

Solution:  

Given: ∠ XYZ = 64°and XY is produced to point P. YQ bisects ∠ ZYP.

To Find: ∠XYQ and reflex ∠QYP. 

YQ bisects ∠ZYP 

∴ ∠ZYQ = ∠QYP   ................. (1) 

 ∵ XY is produced to point P.

∴ PX is a straight line. 

Now, ∠ XYZ + ∠ZYQ + ∠QYP = 180° (linear pair) 

Or,     64° + ∠ZYQ + ∠QYP = 180°

⇒      ∠ZYQ + ∠QYP = 180° - 64°

⇒      ∠ZYQ + ∠ZYQ = 116°                 [Using equation (1) ]

⇒      2∠ZYQ = 116°

⇒      ∠ZYQ = 116°/2

⇒      ∠ZYQ = 58°

∠ZYQ = ∠QYP = 58°

∠XYQ = ∠XYZ + ∠ZYQ 

           = 64° + 58°

           = 122°

∵ ∠QYP = 58°

∴ Reflex ∠QYP = 360° - 58°

                        = 302°

∠XYQ  = 122°, Reflex ∠QYP = 302°

 

Chapter 6. Lines and Angles

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Exercise 6.2

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

Solution: 

x + 50° = 180° (linear pair) 

x = 180° - 50°

x = 130°       ............. (1) 

y = 130° (Vertically oposite angle) ....... (2) 

From equation (1) and (2) 

x = y = 130° (Alternate Interior Angle ) 

∴ AB || CD 

(If alternate interior angle is equal then a pair of lines are parallel.) 

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

Solution:

AB || CD  (Given) ............. (1)

CD || EF  (Given) ............. (2)

From equation (1) and (2) 

AB || EF  

∴ x = z  .......... (3) (Alternate Interior Angle)

y : z = 3 : 7 (Given)

Let y = 3k, z = 7k 

∴ x = z = 7k   From equ. (3) 

AB || CD  (Given)

Now, x + y = 180°  (Sum of interior adjacent angle is 180°)

⇒ 7k + 3k = 180°

⇒ 10k = 180°

⇒ k = 18°

x = 7k 

   = 7 × 18°

 x = 126°

Q3. In Fig. 6.30, if AB || CD, EF ⊥ CD and ∠ GED = 126°, find ∠ AGE, ∠ GEF and ∠ FGE.​

Solution: 

∠GED = 126°

AB || CD and GE is a transversal.  (Given) 

∴ ​ ∠AGE = ∠GED  (Alternate Interior Angle)

 ∴ ​ ∠AGE = 126°

EF ⊥ CD (Given) 

∴ ∠FED = 90°  ............ (1) 

Now, ∠GED = 126°

Or,     ∠GEF + ∠FED = 126°

⇒       ∠GEF +  90°  = 126°      From eqa (1) 

⇒       ∠GEF = 126° - 90°  

⇒       ∠GEF = 36°

 ∠AGE + ∠FGE = 180°  (linear pair) 

⇒  126° + ∠FGE = 180°

⇒  ∠FGE = 180° - 126° 

⇒  ∠FGE = 154°

∠AGE = 126°, ∠GEF = 36°and ∠FGE = 154°

Q4. In Fig. 6.31, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠ QRS.

[Hint : Draw a line parallel to ST through point R.]

Solution: 

Construction: Draw PQ || XY from point R. 

∠ PQR = 110° and ∠ RST = 130°

 PQ || ST  .............. (1) (Given)

PQ || XY .................(2) By construction.

From equa.(1) and (2) we get

ST || XY and SR is a transversal.

∴ ∠ RST + ∠ SRY = 180°  

(Sum of interior Adjacent Angle) 

Or, 130° + ∠ SRY = 180°  

⇒ ∠ SRY = 180° - 130°

⇒ ∠ SRY = 50°

PQ || XY and QR is a transversal 

∴ ∠ PQR = ∠ QRY        (Alternate Interior Angle) 

Or,  ∠ PQR = ∠ QRS +  ∠ SRY 

⇒     110° = ∠ QRS + 50°

⇒     ∠ QRS = 110° - 50°

⇒     ∠ QRS = 60°

Q5. In Fig. 6.32, if AB || CD, ∠ APQ = 50° and ∠ PRD = 127°, find x and y.

Solution: 

Given that:

∠ APQ = 50° and ∠ PRD = 127°

AB || CD and PQ is a transversal.

∴ ∠ PQR = ∠ APQ  (Alternate Interior Angle)

Or, x = 50° 

Similarily,

∠ APR = ∠ PRD  (Alternate Interior Angle)

50° + y = 127°

⇒ y = 127° - 50° 

⇒ y = 77°

x = 50°, y = 77°

Q6. 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.

Solution: 

Given: PQ || RS and AB is incident ray, CD is reflected ray. 

To prove: AB || CD 

Construction:

Draw BM ⊥ PQ and CN ⊥ RS 

Proof: 

 BM ⊥ PQ and CN ⊥ RS 

∴ BM || CM and BC is a transverasal line 

∴ ​∠2 = ∠ 3   ............ (1) (Alternate Interior Angle) 

While we know that 

Angle of incidence = Angle of reflection, where BM and CN are normal. 

∴ ​∠1 = ∠ 2   .............. (2) 

Similarily, 

 ∴ ​∠3 = ∠ 4   .............. (3) 

Using (1) (2) and (3) we get

    ∠1 = ∠ 4  ................ (4) 

Adding equa (1) and (4) 

 ∠1 + ∠2 = ∠ 3 + ∠ 4

 ∠ABC = ∠ BCD  (Alternate Interior Angle) 

Therefore, AB || CD proved  

 

Exercise 6.3

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Q1. In Fig. 6.39, sides QP and RQ of Δ PQR are produced to points S and T respectively. If ∠ SPR = 135° and ∠ PQT = 110°, find ∠ PRQ.

Q2. In Fig. 6.40, ∠ X = 62°, ∠ XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of Δ XYZ, find ∠ OZY and ∠ YOZ.

Q3. In Fig. 6.41, if AB || DE, ∠ BAC = 35° and ∠ CDE = 53°, find ∠ DCE.

Q4. In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠ PRT = 40°, ∠RPT = 95° and ∠ TSQ = 75°, find ∠ SQT.

Q5. In Fig. 6.43, if PQ ⊥ PS, PQ || SR, ∠ SQR = 28° and ∠ QRT = 65°, then find the values of x and y.

Q6. In Fig. 6.44, the side QR of Δ PQR is produced to a point S. If the bisectors of ∠ PQR and ∠ PRS meet at point T, then prove that ∠ QTR = ½ ∠ QPR.