Question

A circle centre O, passes through A, B and C. AT is the tangent to the circle at A. CBT is a straight line. Given that $\angle ABO={68}^o$ and $\angle OBC={20}^o$ the magnitude of $\angle ATB$ is

• A. ${60}^o$
• B. ${64}^o$
• C. ${65}^o$
• D. ${70}^o$
• E. ${66}^o$

Answer 1

The correct option is E ${66}^o$

Consider triangle OAB. Now $OA=OB=radius$. Hence $\angle OAB=\angle OBA$. Therefore $\angle OAB={68}^0$. Now $\angle OAT={90}^0$ ...(point of contact of tangent. Hence $\angle TAB={90}^0-{68}^0={22}^0$. Now CBT is a straight line. Therefore, $\angle CBO+\angle OBA+\angle TBA={180}^0$ or ${20}^0+{68}^0+\angle TBA={180}^0$ or $\angle TBA={92}^0$
Consider triangle ABT, $\angle ATB={180}^0-(\angle ABT+\angle BAT)={180}^0-({92}^0+{22}^0)={180}^0-{114}^0={66}^0$.