A) \[{{90}^{o}}\]
B) \[{{80}^{o}}\]
C) \[{{100}^{o}}\]
D) \[{{120}^{o}}\]
Correct Answer: C
Solution :
Given, \[\angle ABD={{70}^{o}},\angle ADB={{30}^{o}}\] In \[\Delta ADB,\angle DAB+\angle ADB+\angle ABD={{180}^{o}}\] \[\Rightarrow \]\[\angle DAB+{{30}^{o}}+{{70}^{o}}={{180}^{o}}\] \[\Rightarrow \]\[\angle DAB={{180}^{o}}-{{100}^{o}}={{80}^{o}}\] Since, ABCD is a cyclic quadrilateral. \[\therefore \]\[\angle DAB+\angle BCD={{180}^{o}}\] \[\Rightarrow \]\[{{80}^{o}}+\angle BCD={{180}^{o}}\Rightarrow \angle BCD={{100}^{o}}\]
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