Question

plz help me with this question

Answer 1

Refer the attachment for figure

Extend CD such that it meets AE at F

Now, since AB || CD,

=> AB || CF (Since CD is extended and CF is overlapping CD)

=> angle DFE = angle BAE (corresponding angles)

Now in ∆DFE,

angle DFE + angle DEF + angle FDE = 180°

(angle sum property of a triangle)

=> 105 + 25 + angle FDE = 180°

=> angle FDE = 180 - 130

=> angle FDE = 50°

now, angle FDE + x = 180° (linear pair)

=> 50 + x = 180°

=> x = 180 - 50

=> x = 130°


Hope it helps dear friend ☺️

Answer 2

Answer :

Through E, draw EF parallel to AB parallel to CD.

Now, EF parallel to CE is the transversal,

Therefore, angle DCE + angle CEF = 180°
【 consecutive Int. angles 】

=> x° + angle CEF = 180°

=> angle CEF = 180° - x° ----(1)

Again, EF parallel to AB and AE is the transversal,

angle BAE + angle AEF = 180°
【 corres. Int. angles 】

=> 105° + angle AEC + angle CEF = 180°

=> 105° + 25° + ( 180° - x°) = 180° [ using (1) ]

=> 105° + 25° + 180° - x° = 180°

=> 105° + 25° - x° = 0

=> 130° - x° = 0

=> -x° = -130°

=> x° = 130°

=> x = 130

HOPE IT WOULD HELP YOU

PLEASE REFER THE ATTACHMENT AS WELL.