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# ABCD is a rhombus such that $\angle \text{ACB}={{40}^{\circ }}\ then \\angle \text{ADB}$ is ?A. ${{40}^{\circ }}$B. ${{50}^{\circ }}$C. ${{45}^{\circ }}$D. None of these

Last updated date: 16th Jun 2024
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Answer
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Hint:To find the value of $\angle \text{ADB}$, we first find the value of $\angle \text{CBD}$ by using the formula of sum of all the angles equal to ${{180}^{\circ }}$ and after finding the value of angle $\angle \text{CBD}$, we use the alternative angle method where we find the value of $\angle \text{ADB}$.

Complete step by step solution:
Let us draw the diagram and according to the diagram given, we will find the value of the angle $\angle \text{CBD}$. To find the $\angle \text{CBD}$, we first check the triangle $CBO$which contains the angle $\angle OCB$ and $\angle COB$ i.e. ${{40}^{\circ }}$ and ${{90}^{\circ }}$ respectively.

Hence, the value of the angle $\angle \text{CBD}$, we use the formula of the sum of the angle equal to ${{180}^{\circ }}$ where we put the values in the formula as:
$\Rightarrow \angle C+\angle O+\angle B={{180}^{\circ }}$
$\Rightarrow {{40}^{\circ }}+{{90}^{\circ }}+\angle B={{180}^{\circ }}$
$\Rightarrow \angle B={{180}^{\circ }}-{{130}^{\circ }}$
$\Rightarrow \angle B={{50}^{\circ }}$
Now that we have the value of $\angle B$, we can find the value of the $\angle D$ by using the alternate angle method where the angle $CBD$ and $BDA$ are the same. Therefore, the value angle $BDA$ is also given as ${{50}^{\circ }}$.

Note: Rhombus like square have equal sides but the outer angles are not same hence, the outer angle of a rhombus can’t be deemed as ${{90}^{\circ }}$ and similarly student may go wrong if they try to take the outer angles as ${{90}^{\circ }}$ and start their questions with it.