The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O. If `angleDAC=32^(@) and angleAOB=70^(@)`, then `angleDBC` is equal to

Options

A. `24^(@)`
B. `86^(@)`
C. `38^(@)`
D. `32^(@)`

Answer

C

Explanation

Given, `angleAOB=70^(@) and angleDAC=32^(@)`

`angleACB=32^(@)" "[AD||BC and AC` is transversal]
Now, `" "angleAOB+angleBOC=180^(@)" "`[linear pair axiom]
`rArr" "angleBOC+angleBCO+angleOBC=180^(@)` [by angle sum property of a triangle]
`rArr" "110^(@)+32^(@)+angleOBC=180^(@)" "[becauseangleBCO=angleACB=32^(@)]`
`rArr" "angleOBC= 180^(@)-(110^(@)+32^(@))=38^(@)`
`therefore" "angleDBC=angleOBC=38^(@)`