Question

solve this guys important

Answer 1

{For construction work, refer to the Attachment.}

/_BAP + /_EAP = 180°

132° + /_EAP = 180°

/_EAP = 180° - 132°

/_EAP = 48°

Again

/_APC + /_APE = 180°

148° + /_APE = 180°

/_APE = 180° - 148°

/_APE = 32°

Now,

In ∆ EAP,

/_APE + /_EAP + /_PEA = 180° ( Angle Sum Property )

32° + 48° + /_PEA = 180°

80° + /_PEA = 180°

/_PEA = 180° - 80°

/_PEA = 100°

Now,

/_PEA = /_PCD ( Alternate Exterior Angles )

•°•

/_PCD = 100°

x = 100°

Hence,

Option 1) 100° is correct.

^^"

Answer 2

Draw PQ // AB // CD

i ) a + b = 180°

[ Sum of interior angles same side

of the transversal ]

=> a + 132 = 180°

=> a = 180° - 132°

=> a = 48°

ii ) a + y = 148° [ given ]

=> 48° + y = 148°

=> y = 148° - 48°

=> y = 100°

iii ) PQ// CD, PC is the transversal ,

x = y = 100° [corresponding angles ]

Therefore ,

x = 100°

Option ( 1 ) is correct.

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