In the adjoining figure, is a parallelogram in which and

Calculate (i) (ii) (iii) (iv)

In ABCD, and

(i) In ∆AOB,

∠BAO = 35°

∠AOB = ∠COD = 105° …Vertically opposite angels
∴ ​∠ABO = 180° − (35° + 105°) = 40° … Using Angle sum property of Triangle
(ii) ∠ODC and ∠ABO are alternate angles for transversal BD
∴ ∠ODC = ∠ABO = 40°
(iii) ∠ACB = ∠​CAD = 40°° …Alternate angles for transversal AC
(iv) ∠CBD = ∠ABC − ∠ABD ...(1)

∠ABC = 180° − ∠BAD …Adjacent angles are supplementary

​∠ABC = 180° − 75° = 105°
∠CBD = 105° − ∠ABD … as ∠ABD = ∠ABO
∠CBD = 105° − 40° = 65°