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Given that BC = CD, what is the value of x.
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A. \[{{50}^{\circ }}\]
B. \[{{55}^{\circ }}\]
C. \[{{60}^{\circ }}\]
D. \[{{65}^{\circ }}\]

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Last updated date: 24th Jul 2024
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Answer
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Hint: In this problem, we have to find the value of x in the given diagram. We should first analyse the given diagram. We are also given that BC = CD, whose angle will also be equal. We can then use the angle sum property and find and find the value of one of the angles. We can see that in the first circle AEBD is a cyclic quadrilateral, then the opposite angles will be supplementary, we can use these properties and find the value of the required angle.

Complete step by step answer:
Here we have to find the value of x from the given diagram.
seo images

We can see that, in triangle DBC, we are given that \[\angle DCB={{50}^{\circ }}\] and CD = CB, we can now write it as,
\[\Rightarrow \angle CDB=\angle CBD\] …… (1)
By angle sum property, we can write it as,
\[\Rightarrow \angle DCB+\angle DBC+\angle CDB={{180}^{\circ }}\]
From (1), we can write as
\[\begin{align}
  & \Rightarrow 2\angle CDB={{180}^{\circ }}-{{50}^{\circ }} \\
 & \Rightarrow \angle CDB={{65}^{\circ }}.....(2) \\
\end{align}\]
We can see that, in the first circle AEBD is a cyclic quadrilateral, then opposite angles are supplementary.
We can now write it as,
\[\Rightarrow \angle EAB+\angle EDB={{180}^{\circ }}\]
From the diagram, we can also write the above step as,
\[\begin{align}
  & \Rightarrow \angle EAB+{{180}^{\circ }}-\angle CDB={{180}^{\circ }} \\
 & \Rightarrow \angle EAB=\angle CDB \\
\end{align}\]
From (2), we can write it as,
\[\Rightarrow \angle EAB=\angle CDB={{65}^{\circ }}\]
The value of x is \[{{65}^{\circ }}\].
Therefore, the answer is option D. \[{{65}^{\circ }}\]

Note: We should always remember that the supplementary angles are angles which sums up to give \[{{180}^{\circ }}\]. We should also know that if the first circle is a cyclic quadrilateral, then opposite angles are supplementary. We should know that angle sum property is the sum of the triangle equals \[{{180}^{\circ }}\].