PART 1: Find ∠p
Angle which is created by tangent of circle and diameter/radius is 90°.
∠DCO = 90°
We can find ∠ACO by complementary angle property
∠ACO + ∠DCA = 90°
∠ACO + 44° = 90°
∠ACO = 90° - 44°
∠ACO = 46°
Because AOC is iscosceles triangle, the two angles on the leg are the same
∠p = ∠ACO
∠p = 46°
PART 2: Find ∠n
First, find value of smaller ∠AOC with interior angle property of a triangle
the sum of interior angle = 180°
∠AOC + ∠OAC + ∠ACO = 180°
∠AOC + 46° + 46° = 180°
∠AOC + 92° = 180°
∠AOC = 88°
Second, find the bigger ∠AOC with full round angle of 360°
bigger ∠AOC + smaller ∠AOC = 360°
bigger ∠AOC + 88° = 360°
bigger ∠AOC = 272°
Third, find inscribed angle ∠ABC
inscribed angle = 1/2 × central angle
∠ABC = 1/2 × smaller ∠AOC
∠ABC = 1/2 × 88°
∠ABC = 44°
Fourth, find ∠n
sum of interior tetragonal angles is 360°, sum of interior ABCO is 360°
∠n + 272° + 28° + 44° = 360°
∠n + 344° = 360°
∠n = 16°
See my attachment for better understanding
