Question

Question 3 a. Find the values of x and y ZACB = 50 ZDAC = 32° ​

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°