#3 please **Title: Circle Geometry Problems****Diagram Description:**This section features a circle with various points labeled A, B, C, D, and E. Points lie on the circumference and inside the circle, forming angles and line segments that are used in the problems.**Problems 3-6:**3. If \( \overarc{AD} = 125^\circ \) and \( \overarc{BC} = 173^\circ \), find \( \angle BEC \).4. If \( \angle AEB = 35^\circ \) and \( \overarc{AB} = 24^\circ \), find \( \overarc{CD} \).5. If \( \angle CAD = 30^\circ \) and \( \angle BDA = 40^\circ \), find \( \angle CED \).6. If \( \angle CED = 41^\circ \) and \( \angle DAC = 25^\circ \), find \( \overarc{AB} \).**Further Problems 7-10:**These problems reference another circle diagram involving points A, B, C, D, F, and G. These problems involve different sections and angles within this new circle configuration.These exercises help in understanding properties of circles, angles subtended by arcs, and the relationships between inscribed angles and arcs.

Answered: Image /qna-images/answer/e5b5bf87-25e3-430b-b142-eb1c9ceabb08.jpgAim: To find ∠BECSince BD is a straight line.∠BEC = 180° -∠DEC (linear pair) (1)GivenAD⏜ = 125° ; BC⏜ = 173°Theorem to be used: Now we know the angle subtended by the arc or corresponding chord at the center of the circle is double of the angle subtendedby same arc/chord on the circle (on the part of the circle whichis containing then center ). In △BDC∠BDC = 12mBC⏜ (using theorem)∠BDC = 12(173°)∠BDC = 86.5° (2)In △ACD∠ACD = 12mAD⏜ (using theorem)∠ACD = 12(125°)∠ACD = 62.5° (3)HenceIn △ECD∠DEC = 180°- ∠BDC - ∠ACDUsing (2) & (3)∠DEC = 31° (4)Use (4) in (1)∠BEC = 180°-31°∠BEC = 149°