In the figure, tangent $CT$ and chord $BA$ extended meet at $T$. If $\angle CTA = 36^\circ$ and $AT = AC$, find $\angle ABC$.
If length of tangent $CT = 12 \; cm$ and $AT = 8 \; cm$, find the length of chord $AB$.
Given: $\;$ $\angle CTA = 36^\circ$; $\;$ $AT = AC$
$\because \;$ $AT = AC$ $\implies$ $\angle TCA = \angle TAC$ $\;\;$ [angles opposite equal sides are equal]
Now, in $\triangle ATC$,
$\angle TCA + \angle TAC + \angle CTA = 180^\circ$ $\;\;$ [sum of angles of a triangle equal $180^\circ$]
i.e. $\;$ $2 \angle TCA + 36^\circ = 180^\circ$
i.e. $\;$ $2 \angle TCA = 144^\circ$ $\implies$ $\angle TCA = 72^\circ$
Now, $\;$ $\angle TBC = \angle TCA = 72^\circ$ $\;\;$ [angles in alternate segments of the circle]
But points $B$, $A$ and $T$ are collinear.
$\therefore \;$ $\angle ABC = 72^\circ$
From the figure, $\;$ $BT \times AT = CT^2$
[If a chord and a tangent intersect externally, then the product of the lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.]
i.e. $\;$ $BT = \dfrac{CT^2}{AT} = \dfrac{12^2}{8} = 18 \; cm$
But $\;$ $BT = AB + AT$
$\therefore \;$ $AB = BT - AT = 18 - 8 = 10 \; cm$