Question

$\mathrm{ABCD}$ is a rhombus such that $\angle \mathrm{ACB}=40°$ Then $\angle \mathrm{ADB}$ is

• A. $40º$
• B. $45º$
• C. $50º$
• D. $60º$

Answer 1

The correct option is C

$50º$

The explanation for the correct option:

Option (C):$50°$

Solution:

According to the given details, $\square \mathrm{ABCD}$ is a rhombus

$\angle \mathrm{ACB}=40°$

$$\begin{gathered} \because \angle \mathrm{ACB}=40° \\ \Rightarrow \angle \mathrm{OCB}=40° \end{gathered}$$

$\because AD\parallel BC$ (opposite sides of the rhombus are parallel)

$\Rightarrow \angle \mathrm{DAC}=\angle \mathrm{BCA}=40°$ [Alternate interior angles]

$\Rightarrow \angle \mathrm{DAO}=40°$

We know that the diagonals of a rhombus are perpendicular to each other

So,

$\angle \mathrm{AOD}=90°$

We also know that the sum of interior angles of a triangle is $180°.$

Sum of all angles of a triangle $=180°$

$\begin{array}{rcl}& \Rightarrow & \angle \mathrm{AOD}+\angle \mathrm{ADO}+\angle \mathrm{DAO}=180°\\ & \Rightarrow & 90°+\angle \mathrm{ADO}+40°=180°\\ & \Rightarrow & 130°+\angle \mathrm{ADO}=180°\\ & \Rightarrow & \angle \mathrm{ADO}=180°–130°\\ & \Rightarrow & \angle \mathrm{ADO}=50°\\ & \Rightarrow & \angle \mathrm{ADB}=50°\end{array}$

Hence,$\angle \mathrm{ADB}=50°$