In the diagram below circles with centres O and M are given

Question

Please give reasons 

Answer 1

9.1) OC and OD are radii forming isosceles triangle COD and (Ô₁) BOC=110° is the external angle so the two equal angles (Ĉ₃ and D̂₂) are 110/2=55°=D̂₂.

9.2) ∠A=½∠COD (Ô₂)=½(180-110)=35° (angle at the centre is twice the angle at the circumference in the same segment; and angles BOC (Ô₁) and COD (Ô₂) are supplementary).

9.3) ∠B=∠A=35° (angles in the same segment, standing on chord CD).

9.4) ∠ACD=∠ACO (Ĉ₂)+∠OCD (Ĉ₃)=10+55=65°; ∠A=35°, so ∠ADC=180-(65+35)=80° (angles in triangle ACD).

9.5) ∠N=90° (angle in a semicircle is a right angle). ∠ADC=80° (see 9.4). ∠ADB (D̂₁)+∠BDC (D̂₂)=∠ADC, ∠ADB (D̂₁)+55=80, making ∠ADB (D̂₁)=80-55=25°=∠NDQ (D̂₅) (alternate angles at an intersection). ∠Q=90-25=65° (right triangle DNQ).