In the given figure, $\mathrm{BOA}$ is a diameter of a circle and the tangent at a point $\mathrm{P}$ meets $\mathrm{BA}$ produced at $\mathrm{T}$. If $\mathrm{}\mathrm{\angle }\mathrm{P}\mathrm{B}\mathrm{O}=30°$, then find $\mathrm{\angle }\mathrm{P}\mathrm{T}\mathrm{A}$.

A tangent $\mathrm{PQ}$ at a point $\mathrm{P}$ of a circle of radius $5 \mathrm{cm}$ meets a line through the center $\mathrm{O}$ at a point $\mathrm{Q}$ such that $\mathrm{O}\mathrm{Q}=12 \mathrm{c}\mathrm{m}.$ Length $\mathrm{PQ}$ is

From a point $\mathrm{Q}$, the length of the tangent to a circle is $24 \mathrm{c}\mathrm{m}$ and the distance of $\mathrm{Q}$ from the centre is $25 \mathrm{c}\mathrm{m}.$ The radius of the circle is

The length of the tangent from a point $\mathrm{A}$ to a circle of radius $3 \mathrm{c}\mathrm{m},$ is $4 \mathrm{c}\mathrm{m}.$ The distance of $\mathrm{A}$ from the centre of the circle is

If tangents $\mathrm{PA}$ and $\mathrm{PB}$ from a point $\mathrm{P}$ to a circle with center $\mathrm{O}$ are inclined to each other at an angle of $80°$ then $\angle \mathrm{POA}$ is equal to

If $\mathrm{TP}$ and $\mathrm{TQ}$ are two tangents to a circle with centre $\mathrm{O}$ so that $\mathrm{\angle }\mathrm{P}\mathrm{O}\mathrm{Q}=110°,$ then $\mathrm{\angle }\mathrm{P}\mathrm{T}\mathrm{Q}$ is equal to

$\mathrm{PQ}$ is a tangent to a circle with center $\mathrm{O}$ at the point $\mathrm{P}$. If $△\mathrm{O}\mathrm{P}\mathrm{Q}$ is an isosceles triangle, then $\mathrm{\angle }\mathrm{O}\mathrm{Q}\mathrm{P}$ is equal to