Question

If $AB$ is a chord of a circle with center $O,AOC$ is diameter and $AT$ is the tangent at $A$ as shown in Fig. then $\angle BAT=\angle ACB$

• A. True
• B. False

Answer 1

The correct option is A True

Let $\angle ACB=x$ and $\angle BAT=y$

A tangent makes an angle of ${90}^o$ with the radius of a circle.

$\therefore$ $\angle OAT={90}^o$

$\Rightarrow$ $\angle OAB+y={90}^o$

$\Rightarrow$ $\angle CBA={90}^o$ [ Angle inscribe in a semi-circle is ${90}^o$ ]

In $△ABC,$

$\Rightarrow$ $\angle CBA+\angle CAB+\angle ACB={180}^o$ [ Angle sum property of a triangle ]

$\Rightarrow$ ${90}^o+\angle OAB+x={180}^o$

$\Rightarrow$ $\angle OAB+x={90}^o$ ---- ( 2 )

From ( 1 ) and ( 2 ),

$\Rightarrow$ $\angle OAB+y=\angle OAB+x$

$\Rightarrow$ $x=y$

$\therefore$ $\angle ACB=\angle BAT$