In the given figure,$AT$ is a tangent to the circle with center $O$ such that $OT = 4 \, cm$ and $\angle OTA = 30^{\circ}$. Then $AT$ is equal to (in $cm$):

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

$P$ is a point in the exterior of $\odot(O, r)$ and the tangents from $P$ to the circle touch the circle at $X$ and $Y$. Find $r$,if $OP = 30$ and $PY = 24$.

At one end $A$ of a diameter $AB$ of a circle of radius $5 \, cm$,a tangent $XY$ is drawn to the circle. The length of the chord $CD$ parallel to $XY$ and at a distance of $8 \, cm$ from $A$ is (in $cm$):

In $Fig.$,$AB$ and $CD$ are common tangents to two circles of unequal radii. Prove that $AB = CD$.

$\square PQRS$ is a cyclic quadrilateral. If $m \angle P = 30^{\circ}$, then $m \angle R = \dots$ (in $^{\circ}$)