A) \[{{30}^{o}}\]
B) \[{{60}^{o}}\]
C) \[{{45}^{o}}\]
D) \[{{50}^{o}}\]
Correct Answer: A
Solution :
Given that, \[CD||AB\]and \[\angle BAD={{30}^{o}}.\] \[\because \]AOB is the diameter. \[\therefore \]\[\angle ADB={{90}^{o}}\] (Angle in semicircles) Now In \[\Delta ADB,\angle DAB\,+\angle ADB+\angle ABD={{180}^{o}}\] \[{{30}^{o}}+{{90}^{o}}+\angle ABD={{180}^{o}}\Rightarrow \angle ABD={{60}^{o}}\] In cyclic quadrilateral ABCD, we have \[\angle ABD+\angle ACD={{180}^{o}}\] \[\Rightarrow {{60}^{o}}+\angle ACD={{180}^{o}}\Rightarrow \angle ACD={{120}^{o}}\] Also, \[\angle CDA={{30}^{o}}=\angle DAB\](Alternate angles) In \[\Delta CAD,\angle CAD+\angle CDA+\angle ACD={{180}^{o}}\] \[\Rightarrow \]\[\angle CAD+{{30}^{o}}+{{120}^{o}}={{180}^{o}}\Rightarrow \angle CAD={{30}^{o}}\]
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