Question

find all angles of rectangle ABCD​

Answer 1

in triangle AOB

∠OAB+∠OBA+∠AOB=180°

30°+50°+∠AOB=180°

∠AOB=100°

on line DB

∠DOA+∠AOB=180°

∠DOA=80°

in triangle AOD

∠DAO+∠AOD+∠ADO=180°

∠DAO+80°+35°=180°

∠DAO=65°

in ∠ADC

∠ADC=∠ADO+∠ODC

90°=35°+∠ODC

∠ODC=55°

using vertically opposite angle theorem

∠AOB=∠DOC

∠DOC=100°

in triangle DOC

∠ODC+∠DOC+∠OCD=180°

55°+100°+∠OCD=180°

∠OCD=35°

FINALLY THE ANGLES OF RECTANGLE ARE

∠DAB=∠DAO+∠OAB

=65°+30°

=95° but this is not possible for a rectangle

∠ADC=∠ADO+∠ODC

=35°+55°

= 90°

the given figure does not produce a rectangle

hope this helps you

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