In a circle center O (0, r) , two tangents drawn from point A touches circle at point P, Q. If ∠PAQ= 48o then find the ∠APQ? O केन्द्र वाले वृत्त में बिन्दु वाले वृत्त पर बिंदु A से दो स्पर्श रेखा खिची जाती है व्रत को P और Q पर स्पर्श करती है यदि ∠PAQ= 48o तो ∠APQ मान कितना होगा ?

Tangents from an external point to a circle are equal (AP = AQ), making triangle APQ isosceles. The line from the center O to the external point A bisects ∠PAQ, so ∠PAO = 48°/2 = 24°. Since AP = AQ, triangle APQ is isosceles with ∠APQ = ∠AQP = (180° - 48°)/2 = 66°. Alternatively, using right triangle APO: ∠AOP = 180° - 90° - 24° = 66°.