ABCD is a rhombus such that `angleACB=40^(@)`, then `angleADB` is

Options

A. `40^(@)`
B. `45^(@)`
C. `50^(@)`
D. `60^(@)`

Answer

C

Explanation

Given, ABCD is a rhombus such that `angleACB=40^(@)rArrangleOCB=40^(@)`
Since, `AD||BC`
`angleDAC=angleBCA=40^(@)" "["alternate interior angles"]`
Also, `angleAOD=90^(@)`
[diagonals of a rhombus are perpendicular to each other]

We know that, sum of all angles of a triangle ADO is `180^(@)`.
`thereforeangleADO+angleDOA+angleOAD=180^(@)`
`therefore" "angleADO=180^(@) -(40^(@)+90^(@))`
`=180^(@)-130^(@) =50^(@)`
`rArr" " angleADB=50^(@)`