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

The Correct Option is A

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): \(50º\)