Circles

120°

Step 1: We have tangents TA and TB from external point T, with ∠ATB = 60°. Step 2: In quadrilateral TAOB, ∠TAO = ∠TBO = 90° (tangent perpendicular to radius). Step 3: Sum of angles in quadrilateral = 360°, so ∠ATB + ∠AOB + ∠TAO + ∠TBO = 360°. Step 4: Therefore, 60° + ∠AOB + 90° + 90° = 360°, which gives ∠AOB = 120°.

13 cm

Step 1: Let P be the external point, T be the point of contact, and O be the center. Step 2: We have OT = 5 cm (radius) and PT = 12 cm (tangent length). Step 3: Since tangent is perpendicular to radius, triangle POT is a right triangle with right angle at T. Step 4: Using Pythagoras theorem: PO² = PT² + OT² = 12² + 5² = 144 + 25 = 169. Therefore, PO = 13 cm.

10 cm

Step 1: Let the chord be AB with length 16 cm, and let M be the midpoint where perpendicular from center O meets AB. Step 2: Since perpendicular from center bisects the chord, AM = MB = 8 cm. Step 3: Given that OM = 6 cm (distance from center to chord). Step 4: In right triangle OMA, using Pythagoras theorem: OA² = OM² + AM² = 6² + 8² = 36 + 64 = 100. Therefore, radius OA = 10 cm.

100°

Step 1: Let the external point be P and tangents be PA and PB with ∠APB = 80°. Step 2: In quadrilateral PAOB, ∠PAO = ∠PBO = 90° (tangent perpendicular to radius). Step 3: Sum of angles in quadrilateral = 360°, so ∠APB + ∠AOB + ∠PAO + ∠PBO = 360°. Step 4: Therefore, 80° + ∠AOB + 90° + 90° = 360°, which gives ∠AOB = 100°.