M 1 M 2 180

In effigy, l || m, show that ∠1 + ∠2 - ∠3 = 180°.

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Given: fifty || m
To Show: ∠i + ∠2 - ∠3 = 180°
Construction: Through C, draw CF || l || thou

Proof: ∵ fifty || CF by structure and a transversal BC intersects them
∴ ∠i + ∠FCB = 180°
| ∵ Sum of consecutive interior angles on the same side of a transversal is 180°
⇒ ∠1 + ∠2 - ∠FCD = 180° ...(1)
But    ∠FCD = ∠3    ...(ii)
| Alternating interior angles
∴ From (i) and (2),
∠1 + ∠2 - ∠3 = 180°

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

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∵ Lines AB and CD intersect at O
∴ ∠AOC = ∠BOD
| Vertically Opposite Angles
Only ∠BOD = twoscore°    ...(i) | Given
∴ ∠AOC = 40°    ...(2)
Now, ∠AOC + ∠BOE = 70°
⇒ 40° + ∠BOE = 70° | Using (ii)
⇒ ∠BOE = lxx° - 40°
⇒ ∠BOE = 30°
Once more,
Reflex ∠COE
= ∠COD + ∠BOD + ∠BOE
= ∠COD + 40° + 30°
| Using (1) and (2)
= 180° + twoscore° + 30°
| ∵ Ray OA stands on line CD
|∴ ∠AOC + ∠AOD = 180° (Linear Pair Axiom) ⇒ ∠COD = 180°
= 250°.

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In figure, lines XY and MN intersect at O. If ∠POY = xc° and a : b = ii : 3, find c.

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Ray OP stands on  line Xy

                             | Liner Pair Axiom

            a + b = xc⁰ .....(1)

                 a : 2 = two : 3

              a = 2k
                   b = 3k

Putting the values of a and b in (1), we get

           2k + 3k = 90⁰

∵ Ray OX stands on line MN
∴ ∠XOM + ∠XON = 180°
| Linear Pair Axiom
⇒    b + c = 180°
⇒    54° + c = 180° | Using (2)
⇒    c = 180° - 54°
⇒    c = 126°.

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In effigy, ∠PQR = ∠PRQ, and so prove that ∠PQS = ∠PRT.

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∵ Ray QP stands on line ST
∴ ∠PQS + ∠PQR = 180°    ...(1)
| Linear Pair Axiom
∵ Ray RP stands on line ST
∴ ∠PRQ + ∠PRT = 180°    ...(ii)
| Linear Pair Axiom
From (1) and (2), we obtain
∠PQS + ∠PQR = ∠PRQ + ∠PRT
⇒    ∠PQS = ∠PRT.
| ∵ ∠PQR = ∠PRQ (Given)

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In figure, if x + y = w + z, and then show that AOB is a line.

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∴ AOB is a line.

| If the sum of two side by side angles is 180°, then the non-common arms of the angles form a line.

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In figure, POQ is a line. Ray OR is perpendicular to line PQ. Bone is some other ray lying betwixt rays OP and OR. Prove that

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∵ Ray OR is perpendicular to line PQ.
∴ ∠QOR = ∠POR = ninety°    ...(1)
∠QOS = ∠QOR + ∠ROS    ...(2)
∠POS = ∠POR - ∠ROS    ...(3)
From (2) and (3),
∴ ∠QOS - ∠POS = (∠QOR - ∠POR) + two∠ROS = two∠ROS | Using (ane)

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M 1 M 2 180,

Source: https://www.zigya.com/study/book?class=9&board=cbse&subject=Mathematics&book=Mathematics&chapter=Lines+and+Angles&q_type=&q_topic=&q_category=&question_id=MAEN9056790

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