A) \[{{45}^{o}}\]
B) \[{{58}^{o}}\]
C) \[{{60}^{o}}\]
D) \[{{40}^{o}}\]
Correct Answer: D
Solution :
Given, ABCD is a cyclic quadrilateral, \[\angle ADC={{130}^{o}}\] \[\therefore \]\[\angle ADC+\angle ABC={{180}^{o}}\] \[\Rightarrow \]\[\angle ABC={{50}^{o}}\] Since, AB is a diameter of circle \[\therefore \]\[\angle ACB={{90}^{o}}\] Now, in \[\Delta ACB,\] \[\angle ABC+\angle ACB+\angle BAC={{180}^{o}}\] \[\Rightarrow \]\[{{50}^{o}}+{{90}^{o}}+\angle BAC={{180}^{o}}\Rightarrow \angle BAC={{40}^{o}}\]You need to login to perform this action.
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