Question

If ABCD is a rhombus such that $\angle ACB={40}^{\circ }$, then find $\angle ADB$.

• A. ${50}^{\circ }$
• B. ${60}^{\circ }$
• C. ${45}^{\circ }$
• D. ${40}^{\circ }$

Answer 1

The correct option is A ${50}^{\circ }$

Given: ABCD is a rhombus such that $\angle ACB={40}^{\circ }$ and $\angle OCB={40}^{\circ }$
Since $AD||BC$,
$\angle DAO=\angle BCA={40}^{\circ }$ (alternate interior angle)
$\angle AOD={90}^{\circ }$ (diagonals of rhombus are perpendicular)
WKT, Sum of all angles of triangle ADO is ${180}^{\circ }$
$\angle ADO+\angle DOA+\angle OAD={180}^{\circ }$
$\angle ADO={50}^{\circ }$