Math, asked by maitreyapatni25, 10 months ago

In the given figure, AB || CD || EF intersected by a

transversal PQ at X, Y and Z respectively. Find the

∠BXY, ∠DYZ and ∠YZE.​

Attachments:

Answers

Answered by amitnrw
3

Given:  AB || CD || EF intersected by a  transversal PQ at X, Y and Z respectively  ∠AXP = 8a , ∠FZY = 7a

To find : ∠BXY, ∠DYZ and ∠YZE.​

Solution:

∠BXY = ∠AXP  ( vertically opposite angle)

∠AXP = 8a

=> ∠BXY = 8a

∠BXZ  +   ∠FZP = 180°    ( angles created in between the parallel lines AB & EF)

∠BXZ  = ∠BXY = 8a

∠FZP  = 7a

=> 8a + 7a = 180°

=> 15a = 180°

=> a = 12°

∠BXY = 8a = 8 * 12 = 96°

∠BXY = 96°

∠DYZ = ∠FZP  = 7a = 84°

∠FZP  =  84°

∠YZE + ∠FZY  = 180°  Straight line

∠FZY = ∠FZP   = 84°

=> ∠YZE + 84° = 180°

=> ∠YZE = 96°

Learn more:

7. If a transversal intersects two lines parallel lines then prove that ...

https://brainly.in/question/18440350

Line AB || line CD || line EF and line QP is their transversal. If y:z = 3 ...

https://brainly.in/question/4557443

Answered by ssshiva880
0

Answer:

Solution

Step-by-step explanation:

∠BXY = ∠AXP  ( vertically opposite angle)

∠AXP = 8a

=> ∠BXY = 8a

∠BXZ  +   ∠FZP = 180°    ( angles created in between the parallel lines AB & EF)

∠BXZ  = ∠BXY = 8a

∠FZP  = 7a

=> 8a + 7a = 180°

=> 15a = 180°

=> a = 12°

∠BXY = 8a = 8 * 12 = 96°

∠BXY = 96°

∠DYZ = ∠FZP  = 7a = 84°

∠FZP  =  84°

∠YZE + ∠FZY  = 180°  Straight line

∠FZY = ∠FZP   = 84°

=> ∠YZE + 84° = 180°

=> ∠YZE = 96°

Similar questions