Question

ABCD is a cyclic trapezium and AB || CD. If AB be the diameter of the circle and \[\angle CAB={{30}^{\circ }}\], then the value of \[\angle ADC=\]
(a)\[{{110}^{\circ }}\]
(b)\[{{115}^{\circ }}\]
(c)\[{{120}^{\circ }}\]
(d)\[{{125}^{\circ }}\]

Answer 1

Hint: Draw a cyclic trapezium with the given conditions. An angle inscribed in a semi – circle is \[{{90}^{\circ }}\]. Find the angle of \[{{1}^{st}}\]part. Similarly, the sum of opposite sides of a cyclic quadrilateral is \[{{180}^{\circ }}\].

Complete step-by-step answer:
Consider the figure given.

It is said that ABCD is a cyclic trapezium.
We can say that AB is parallel to CD\[\Rightarrow \]AB || CD.
Given that, \[\angle CAB={{30}^{\circ }}\].
As AB || CD and by angle inscribed in a semi – circle, we can say that, \[\angle ACB={{90}^{\circ }}\].
Or an angle inscribed in a semi – circle is always \[{{90}^{\circ }}\].
Now let us consider, \[\vartriangle ACB\].
We know that the sum of the angles of a triangle is \[{{180}^{\circ }}\].
We know, \[\angle CAB={{30}^{\circ }}\], \[\angle ACB={{90}^{\circ }}\]. Let us find \[\angle CBA\].
In \[\vartriangle ABC\],
\[\begin{align}
  & \angle CAB+\angle ACB+\angle CBA={{180}^{\circ }} \\
 & {{30}^{\circ }}+{{90}^{\circ }}+\angle CBA={{180}^{\circ }} \\
 & \angle CBA={{180}^{\circ }}-{{90}^{\circ }}-{{30}^{\circ }} \\
 & \angle CBA={{60}^{\circ }} \\
\end{align}\]
We know that the sum of opposite angles of a cyclic quadrilateral is \[{{180}^{\circ }}\].
Thus from cyclic quadrilateral ABCD, we can say that,
\[\begin{align}
  & \angle ADC+\angle ABC={{180}^{\circ }} \\
 & \Rightarrow \angle ADC+{{60}^{\circ }}={{180}^{\circ }} \\
 & \angle ADC={{180}^{\circ }}-{{60}^{\circ }}={{120}^{\circ }} \\
\end{align}\]
Thus we got the required angle \[\angle ADC\] as \[{{120}^{\circ }}\].
\[\therefore \]Option (c) is the correct answer.
Note: We have used two important properties here. Remember the properties that is the angle inscribed in a semi – circle is \[{{90}^{\circ }}\] and the sum of opposite angles of a cyclic quadrilateral is \[{{180}^{\circ }}\]. We have solved this question with these 2 theorems.