Answer:
See text below or attached figure
Step-by-step explanation:
Given arcs
AC=70
CR=18
therefore AR = 88
RB=80
BE=130
therefor EA = 360-(70+18+80+130) = 360-298 = 62
angles will be denoted (1) for angle 1, etc.
We ASSUME
ARD is a straight line
PFRB is a straight line
FCE is a straight line
Using inscribed angle theorem, angles subtended by chords/arcs equal to half the arc central angle.
Therefore
(4)=80/2=40
(13)=130/2=65
(12)=62/2=31
(11)=70/2=35
(5) = (70+18)/2 = 44
Consider triangle AEG,
(7)=(13)+(11)=65+35=100 [exterior angle]
Consider triangle EGB,
(10)=180-100-31 = 49 [sum of angles of a triangle]
Consider triangle AEH,
(3) = 180-(4)-(13)-(11) = 180-40-65-35 = 40 [sum of angles of a triangle]
From cyclic quadrilateral ARBE,
ARB+AEB=180 =>
ARB=180-AEB=180-(35+49) = 96
By the intersecting secants theorem,
(2) = (130-18)/2 = 56 [secants FE, FB]
(1) = (130+62 - (18+70))/2 = 104/2 = 52 [secants PA,PB]
(8) = (130+62 -80)/2 = 112/2 = 56
ARD is straight line (see assumptions above)
(9) = 180-96 = 84 [sum of angles on a line]
ARP = (9) = 84 [vertically opposite angles]
Consider triangle ARP
(14) = 180-52-84 = 44
Consider tangent PA
(15) = 180-(44+40+65) = 31 [sum of angles of a triangle]
Consider triangle ABD
(6) = 180 - (40+44+56) = 40 [sum of angles of a triangle]
This completes the search for all sixteen angles, as shown in the diagram, or in the text above.
