**1. ** We are given that ∠AOC + ∠BOE = 70^{0} and ∠BOD = 40^{0}.

We need to find ∠BOE and reflex ∠COE

From the given figure, we can conclude that ∠COB and ∠COE form a linear pair. We know that sum of the angles of a linear pair is 180^{0}.

∴ ∠COB + ∠COE = 180^{0}

∴ ∠COB = ∠AOC + ∠BOE, or

∴ ∠AOC + ∠BOE + ∠COE = 180^{0}

⇒ 70^{0} + ∠COE = 180^{0}

⇒ ∠COE = 180^{0} – 70^{0}

= 110^{0}.

Reflex ∠COE = 360^{0 }– ∠COE

= 360^{0} – 110^{0}

= 250^{0}.

∠AOC = ∠BOD (Vertically opposite angles), or

∠BOD + ∠BOE = 70^{0}.

But, we are given that ∠BOD = 40^{0}.

40^{0} + ∠BOE = 70^{0} ∠BOE = 70^{0} – 40^{0}

= 30^{0}.

Therefore, we can conclude that Reflex ∠COE = 250^{0} and ∠BOE = 30^{0}.

**2. ** We are given that ∠POY = 90^{0} and a : b = 2 : 3 .

We need find the value of c in the given figure.

Let a be equal to 2x and b be equal to 3x.

∵ a + b = 90^{0} ⇒ 2x + 3x = 90^{0} ⇒ 5x = 90^{0}

⇒ x = 18^{0}

Therefore b = 3 × 18^{0} = 54^{0}

Now b + c = 180^{0} [Linear pair]

⇒ 54^{0 }+ c = 180^{0}

⇒ c = 180^{0} – 54^{0} = 126^{0}

**3. ** We need to prove that ∠PQS = ∠PRT

We are given that ∠PQR = ∠PRQ

From the given figure, we can conclude that ∠PQS and ∠PQR, and ∠PRS and ∠PRT form a linear pair.

We know that sum of the angles of a linear pair is 180^{0}.

∴ ∠PQS + ∠PQR = 180^{0}, and (i)

∠PRQ + ∠PRT = 180^{0}. (ii)

From equations (i) and (a), we can conclude that

∠PQS + ∠PQR = ∠PRQ + ∠PRT

But, ∠PQR = ∠PRQ.

∴ ∠PQS = ∠PRT

Therefore, the desired result is proved.

**4. ** We need to prove that AOB is a line.

We are given that x + y = w + z.

We know that the sum of all the angles around a fixed point is 360^{0}. Thus, we can conclude that ∠A0C + ∠BOC + ∠AOD + ∠BOD = 360^{0} or y + x + z + w = 360^{0}.

But, x + y = w + z (Given).

2 (y + x) = 360^{0}.

y + x = 180^{0}.

From the given figure, we can conclude that y and x form a linear pair.

We know that if a ray stands on a straight line, then the sum of the angles of linear pair formed by the ray with respect to the line is 180^{0}.

y + x = 180^{0}.

Therefore, we can conclude that AOB is a line.

We are given that OR is perpendicular to PQ, or

∠QOR = 90^{0}.

From the given figure, we can conclude that ∠POR and ∠QOR form a linear pair. We know that sum of the angles of a linear pair is 180^{0}.

∴ ∠POR + ∠QOR = 180^{0}, or

∠POR = 90^{0}.

From the figure, we can conclude that ∠POR = ∠POS + ∠ROS

⇒ ∠POS + ∠ROS = 90^{0}, or

∠ROS = 90^{0} – ∠POS (i)

From the given figure, we can conclude that ∠QOS and ∠POS form a linear pair.

We know that sum of the angles of a linear pair is 180^{0}.

∠QOS + ∠POS = 180^{0}, or

Therefore, the desired result is proved.

**6. ** We are given that ∠XYZ = 64^{0}, XY is produced to P and YQ bisects ∠ZYP

We can conclude the given below figure for the given situation:

We need to find ∠XYQ and reflex ∠QYP

From the given figure, we can conclude that ∠XYZ and ∠ZYP form a linear pair. We know that sum of the angles of a linear pair is 180^{0}.

∠XYZ + ∠ZYP = 180^{0}.

But ∠XYZ = 64^{0}.

⇒ 64 + ∠ZYP =180^{0}

⇒ ∠ZYP =116^{0}.

Ray YQ bisects ∠ZYP , or

∠XYQ = ∠QYZ + ∠XYZ

= 58^{0}+ 64^{0} = 122^{0}.

Reflex ∠QYP = 360^{0} – ∠QYP

= 360^{0} – 58^{0}

= 302^{0}.

Therefore, we can conclude that ∠XYQ = 122^{0} and Reflex ∠QYP = 302^{0}.

**7. ** We need to find the value of x and y in the figure given below and then prove that

AB || CD.

From the figure, we can conclude that

y = 130 (Vertically opposite angles), and

x and 50^{0} form a pair of linear pair.

We know that the sum of linear pair of angles is 180^{0}.

x + 50^{0} = 180^{0}

x = 130^{0}.

x = y = 130^{0}.

From the figure, we can conclude that x and y form a pair of alternate interior angles corresponding to the lines AB and CD. Therefore, we can conclude that x = 130^{0}, y = 130^{0} and AB || CD.

**8.** We are given that AB || CD , CD ⊥ EF and y : z = 3 : 7.

We need to find the value of x in the figure given below.

We know that lines parallel to the same line are also parallel to each other.

We can conclude that AB || CD || EF.

Let y = 3a and z = 7 a.

We know that angles on same side of a transversal are supplementary.

x + y = 180^{0}.

x = z (Alternate interior angles)

z + y = 180^{0}, or

7 a + 3a = 180^{0} ⇒ 10a =180^{0}

a = 180^{0}.

z = 7 a = 126^{0}

y = 3a = 54^{0}.

Now x + 54 = 180^{0}

x = 126.

Therefore, we can conclude that x = 126.

**9. ** We are given that AB || CD , EF ⊥ CD and ∠GED = 126^{0}.

We need to find the value of ∠AGS, ∠GEF and ∠FGE in the figure given below.

∠GED = 126^{0}

∠GED = ∠FED + ∠GEF.

But, ∠FED = 90^{0}.

126^{0} = 90^{0} + ∠GEF ⇒ ∠GEF = 36^{0}.

∠AGE = ∠GED (Alternate angles)

∠AGE =126^{0}.

From the given figure, we can conclude that ∠FED and ∠FEC form a linear pair. We know that sum of the angles of a linear pair is 180^{0}.

∠FED + ∠FEC = 180^{0} ⇒ 90^{0 }+ ∠FEC = 180^{0}

∠FEC = 90^{0}

But ∠FEC = ∠GEF + ZGEC

90^{0} = 36^{0} + ∠GEC ⇒ ∠GEC = 54^{0}.

∠GEC = ∠FGE = 54^{0} (Alternate interior angles).

Therefore, we can conclude that ∠AGE = 126^{0}, ∠GEF = 36^{0} and ∠FGE = 54^{0}.

**10. ** We are given that PQ || ST, ∠PQR = 110^{0} and ∠RST = 130^{0}.

We need to find the value of ∠QRS in the 110^{0} figure.

We need to draw a line RX that is parallel to the line ST, to get Thus, we have ST || RX.

We know that lines parallel to the same line are also parallel to each other. We can conclude that PQ || ST || RX .

∠PQR = ∠QRX (Alternate interior angles), or

∠QRX =110^{0}.

We know that angles on same side of a transversal are supplementary.

∠RST + ∠SRX = 180^{0} ⇒ 130^{0} + ∠SRX =180^{0}

⇒ ∠SRX =180^{0} – 130^{0} = 50^{0}.

From the figure, we can conclude that

∠QRX = ∠SRX + ∠QRS ⇒ 110^{0} = 50^{0} + ∠QRS

⇒ ∠QRS = 60^{0}.

Therefore, we can conclude that ∠QRS = 60^{0}.

**11.** We are given that AB || CD, ∠APQ = 50^{0} and ∠PRD = 127^{0}.

We need to find the value of x and y in the figure.

∠APQ = x = 50^{0}. (Alternate interior angles)

∠PRD = ∠APR = 127^{0}. (Alternate interior angles)

∠APR = ∠QPR + ∠APQ.

127^{0} = y + 50^{0} y = 77^{0}.

Therefore, we can conclude that x = 50^{0} and y = 77^{0}.

**12. ** We are given that PQ and RS are two mirrors that are parallel to each other.

We need to prove that AB || CD in the figure.

Let us draw lines BX and CY that are parallel to each other, to get

We know that according to the laws of reflection

∠ABX = ∠CBX and ∠BCY = ∠DCY

∠BCY = ∠CBX (Alternate interior angles)

We can conclude that ∠ABX = ∠CBX = ∠BCY = ∠DCY.

From the figure, we can conclude that

∠ABC = ∠ABX + ∠CBX, and

∠DCB = ∠BCY + ∠DCY .

Therefore, we can conclude that ∠ABC = ∠DCB.

From the figure, we can conclude that ∠ABC and ∠DCB form a pair of alternate interior angles corresponding to the lines AB and CD, and transversal BC.

Therefore, we can conclude that AB || CD.

**13. **We are given that ∠SPR = 135^{0} and ∠PQT = 110^{0}.

We need to find the value of ∠PRQ in the figure given below. From the figure, we can conclude that

∠SPR and ∠RPQ, and ∠SPR and ∠RPQ form a linear pair.

We know that the sum of angles of a linear pair is 180^{0}

∠SPR + ∠RPQ = 180^{0}, and

∠PQT + ∠PQR = 180^{0}.

135^{0 }+ ∠RPQ = 180^{0}, and

110^{0} + ∠PQR = 180^{0}, or

∠RPQ = 45^{0}, and

∠PQR = 70^{0}.

From the figure, we can conclude that

∠PQR + ∠RPQ + ∠PRQ = 180^{0}. (Angle sum property)

⇒ 70^{0} + 45^{0} + ∠PRQ = 180^{0} ⇒ 115^{0} + ∠PRQ = 180^{0}

⇒ ∠PRQ = 65^{0}.

Therefore, we can conclude that ∠PRQ = 65^{0}.

**14. ** We are given that ∠X = 62^{0}, ∠XYZ = 54^{0} and YO and ZO are bisectors of ∠XYZ and ∠XZY , respectively. We need to find ∠OZY and ∠YOZ in the figure.

From the figure, we can conclude that in Δ XYZ

∠X + ∠XYZ + ∠XZY = 180^{0}. (Angle sum property)

⇒ 62^{0} + 54^{0} + ∠XZY = 180^{0} ⇒ 116^{0} + ∠XZY = 180^{0}

⇒ ∠XZY = 64^{0}.

We are given that OY and OZ are the bisectors of ∠XYZ and ∠XZY , respectively.

From the figure, we can conclude that in Δ OYZ

∠OYZ + ∠OZY + ∠YOZ = 180^{0}. (Angle sum property)

27^{0} + 32^{0} + ∠YOZ = 180^{0} ⇒ 59^{0} + ∠YOZ =180^{0}

⇒ ∠YOZ = 121^{0}.

Therefore, we can conclude that ∠YOZ = 121^{0} and ∠OZY = 32^{0}.

**15. **We are given that AB || DE, ∠BAC = 35^{0} and ∠CDE = 53^{0}.

We need to find the value of ∠DCE in the figure given below.

From the figure, we can conclude that

∠BAC = 35^{0} and ∠CDE = 53^{0}.

∠BAC = ∠CED = 35^{0} (Alternate interior angles)

From the figure, we can conclude that in Δ DCE

∠DCE + ∠CED + ∠CDE = 180^{0}. (Angle sum property)

∠DCE + 35^{0} + 53^{0} = 180^{0} ⇒ ∠DCE + 88^{0} = 180^{0}

⇒ ∠DCE = 92^{0}.

Therefore, we can conclude that ∠DCE = 92^{0}.

**16. ** We are given that ∠PRT = 40^{0} ∠RPT = 95^{0} and ∠TSQ = 75^{0}.

We need to find the value of ∠SQT in the figure.

From the figure, we can conclude that in Δ RTP

∠PRT + ∠RTP + ∠RPT = 180^{0}. (Angle sum property)

40^{0} + ∠RTP + 95^{0} = 180^{0} ⇒ ∠RTP + 135^{0} = 180^{0}

⇒ ∠RTP = 45^{0}.

From the figure, we can conclude that

∠RTP = ∠STQ = 45^{0}. (Vertically opposite angles)

From the figure, we can conclude that in Δ STQ

∠SQT + ∠STQ + ∠TSQ = 180^{0}. (Angle sum property)

∠SQT + 45^{0} + 75^{0} = 180^{0} ⇒ ∠SQT + 120^{0} = 180^{0}

⇒ ∠SQT = 60^{0}.

Therefore, we can conclude that ∠SQT = 60^{0}.

**17. **We are given that PQ ⊥ PS , PQ || SR, ∠SQR = 28^{0} and ∠QRT = 65^{0}.

We need to find the values of x and y in the figure.

We know that “If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.”

From the figure, we can conclude that

∠SQR + ∠QSR = ∠QRT , or

28^{0} + ∠QSR = 65^{0} ⇒ ∠QSR = 37^{0}.

From the figure, we can conclude that

x = ∠QSR = 37^{0} (Alternate interior angles)

From the figure, we can conclude that Δ PQS

∠PQS + ∠QSP + ∠QPS = 180^{0}. (Angle sum property)

∠QPS = 90^{0} (PQ ⊥ PS)

x + y + 90^{0} = 180^{0} ⇒ x + 37^{0}+ 90^{0} = 180^{0}

x + 127^{0} = 180^{0} ⇒ x = 53^{0}.

Therefore, we can conclude that x = 53^{0} and y = 37^{0}.

**18. **We need to prove that in the figure given below.

We know that “If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.”

From the figure, we can conclude that in Δ QTR , ∠TRS is an exterior angle

∠QTR + ∠TQR = ∠TRS, or

∠QTR = ∠TRS – ∠TQR …(i)

From the figure, we can conclude that in Δ QTR , ∠TRS is an exterior angle

∠QPR + ∠PQR = ∠PRS.

We are given that QT and RT are angle bisectors of ∠PQR and ∠PRS.

∠QPR + 2∠TQR = 2∠TRS

∠QPR = 2 (∠TRS – ∠TQR).

We need to substitute equation (i) in the above equation, to get

∠QPR = 2∠QTR, or

Therefore, we can conclude that the desired result is proved.