# In fig., if TP and TQ are the two tangents to a circle with centre O so that

Question:

In fig., if TP and TQ are the two tangents to a circle with centre O so that $\angle \mathrm{POQ}=110^{\circ}$, then $\angle \mathrm{PTO}$ is equal to -

(A) 60°

(B) 70°

(C) 80° P Q T O 110°

(D) 90°

Solution:

$\mathrm{TQ}$ and $\mathrm{TP}$ are tangents to a circle with centre $\mathrm{O}$ and $\angle \mathrm{POQ}=110^{\circ}$

$\therefore \mathrm{OP} \perp \mathrm{PT}$ and $\mathrm{OQ} \perp \mathrm{QT}$

$\Rightarrow \angle \mathrm{OPT}=90^{\circ}$ and $\angle \mathrm{OOT}=90^{\circ}$

Now, in the quadrilateral TPOQ, we get

$\therefore \mathrm{PTQ}+90^{\circ}+110^{\circ}+90^{\circ}=360^{\circ}$

[Angle sum property of a quadrilateral]

$\Rightarrow \angle \mathrm{PTQ}+290^{\circ}=360^{\circ}$

$\Rightarrow \angle \mathrm{PTQ}=360^{\circ}-290^{\circ}=70^{\circ}$

Hence, the correct option is (B)