Question

plz answer this question​

Answer 1

Answer:

Given, ABCD is a parallelogram having ∠BAO = 35°, ∠DAO = 40° and ∠COD = 105°

Now, ∠COD = ∠AOB =  105° [vertically opposite angles]

In ΔAOB, by angle sum property of triangle,  

⇒ ∠AOB + ∠OAB + ∠ABO =  180°

⇒ 105° + 35° + ∠ABO =  180°

⇒ ∠ABO =  40°

Again, adjacent angles of a parallelogram are supplementary.

⇒ ∠DAB + ∠ABC =  180°

⇒ ∠DAO + ∠OAB + ∠ABO + ∠CBO  =  180°

⇒ 40° + 35° + 40° + ∠CBO  =  180°

⇒ ∠CBO  = ∠CBD =  180° - 115° = 65°  

⇒ ∠CBD =  65°  

In ΔABC, by angle sum property of triangle,  

⇒ ∠CAB + ∠ABC + ∠ACB =  180°

⇒ 35° + ∠ABO + ∠CBO + ∠ACB =  180°

⇒ 35° + 40° + 65° + ∠ACB =  180°

⇒ ∠ACB = 180° - 140° = 40°  

⇒ ∠ACB = 40°  

Now, opposite angles of a parallelogram are equal

⇒ ∠A =∠C  

⇒ ∠C  =  75°

On applying angle sum property of triangle in BCD, we get

⇒ ∠C  + ∠CBD + ∠CDB =  180°

⇒ 75°  + 65° + ∠CDB =  180°

⇒ ∠CDB =  40°

or ∠ODC =  40°