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.
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.