question_answer

If O is the centre of a circle, AOC is its diameter and B is a point an the circle such that\[\angle ACB={{50}^{o}}\]. If AT is the tangent to the circle at the point A, then \[\angle BAT=\]

A) \[{{40}^{o}}\]                     

B)        \[{{50}^{o}}\]                                 

C)        \[{{60}^{o}}\]                     

D)        \[{{65}^{o}}\]        

Correct Answer: B

Solution :

\[\angle ABC={{90}^{o}}\]       [Angle in a semicircle ] In \[\Delta ABC,\] we have                              \[\angle ABC+\angle CAB+\angle ABC={{180}^{o}}\] \[\Rightarrow \]            \[{{50}^{o}}+\angle CAB+{{90}^{o}}={{180}^{o}}\] \[\Rightarrow \]            \[\angle CAB={{40}^{o}}\] Now, \[\angle CAT={{90}^{o}}\Rightarrow \angle CAB+\angle BAT={{90}^{o}}\] \[\Rightarrow \] \[{{40}^{o}}+\angle BAT={{90}^{o}}\Rightarrow \angle BAT={{50}^{o}}\]