In $Fig.$,$PQ$ is a chord of a circle and $PT$ is the tangent at $P$ such that $\angle QPT = 60^{\circ}$. Then $\angle PRQ$ is equal to (in $^{\circ}$)

$A$ circle with centre $O$ touches sides $\overline{AB}$,$\overline{BC}$,$\overline{CD}$ and $\overline{DA}$ of quadrilateral $ABCD$ at points $P, Q, R$ and $S$ respectively. Prove that $m \angle AOB + m \angle COD = 180^{\circ}$ and $m \angle AOD + m \angle BOC = 180^{\circ}$.

$P$ lies in the exterior of $\odot(O, 8)$ such that $OP = 17$. Two tangents are drawn to the circle which touch the circle at $A$ and $B$. Find $AB$.

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 $\angle BAT = \angle ACB$.

If $a, b, c$ are the sides of a right triangle where $c$ is the hypotenuse,prove that the radius $r$ of the incircle of the triangle is given by $r = \frac{a + b - c}{2}$.

$\overline{ PA }$ and $\overline{ PB }$ are the tangents to $\odot( O , r)$ drawn from a point $P$ outside a circle. If $m \angle APB = 65^{\circ}$,then $m \angle AOB = \ldots \ldots \ldots . .$ (in $^{\circ}$)