# 3.   If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of $80 \degree$, then $\angle$ POA is equal to      (A) 50°      (B) 60°      (C) 70°      (D) 80°

M manish

The correct option is (A)

It is given that,  tangent PA and PB from point P inclined at $\angle APB = 80^0$
In triangle $\Delta$OAP and $\Delta$OBP
$\angle OAP = \angle OBP = 90$
OA =OB (radii of the circle)
PA = PB (tangents of the circle)

Therefore, by  SAS congruence
$\therefore \Delta OAP \cong \Delta OBP$

By CPCT, $\angle OPA = \angle OPB$
Now, $\angle$OPA = 80/4 = 40

In $\Delta$ PAO,
$\angle P + \angle A + \angle O = 180$
$\angle O = 180 - 130$
= 50

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