Question

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to

• A. 50°
• B. 60°
• C. 70°
• D. 80°

Answer 1

Answer: (A)

50°

Answer 2

It is given that $PA$ and $PB$ are tangents.

Therefore, the radius drawn to these tangents will be perpendicular to the tangents.
Thus, $OA ⊥ PA$ and $OB ⊥ PB$
$∠OBP = 90º$
$∠OAP = 90º$
In $AOBP$,
Sum of all interior angles $= 360º$
$∠OAP + ∠APB +∠PBO + ∠BOA = 360º$
$90º + 80º +90º + BOA = 360º$
$∠BOA = 100º$
In $ΔOPB$ and $ΔOPA$,
$AP = BP$ (Tangents from a point)
$OA = OB$ (Radii of the circle)
$OP = OP$ (Common side)
Therefore, $ΔOPB ≅ ΔOPA$ (SSS congruence criterion)
$A ↔ B, P ↔ P, O ↔ O$
And thus, $∠POB = ∠POA$
$∠POA = \frac 12∠AOB$

$∠POA = \frac {100º}{2}$
$∠POA=$ $50º$

Hence, the correct option is (A): $50º$