+4622 In the figure, point \(A\) is the center of the circle, the measure of angle \(RAS\) is 74 degrees, and the measure of angle \(RTB\) is 28 degrees. What is the measure of minor arc \(BR\) , in degrees?
+129899 Call the other intersection of BT and the circle, U
So angles
RAB + RAS + SAU = 180
RAB + 74 + SAU = 180
RAB + SAU = 106
SAU = 106 - RAB
And we have that
m RTB = (1/2) (minor arc RB - minor arc SU)
And since RAB and SAU are central angles, their measures = the measure of their minor arcs...so..
28 = (1/2) ( RAB - SAU )
56 = ( RAB - SAU)
56 = (RAB- (106 - RAB) )
56 = 2RAB - 106
162 = 2RAB
81° = RAB = minor arc RB = minor arc BR
+9481 
https://i.imgur.com/sGRRoeR.png
Since AR and AS are radii of circle A, they are the same length. So triangle RAS is isosceles, and base angles ARS and ASR have the same measure.
74° + m∠ASR + m∠ARS = 180°
74° + m∠ASR + m∠ASR = 180°
74° + 2m∠ASR = 180°
2m∠ASR = 106°
m∠ASR = 53°
m∠AST = 180° - m∠ASR = 180° - 53° = 127°
m∠SAT = 180° - 28° - 127° = 25°
m∠BAR = 180° - 74° - 25° = 81°
measure of arc BR = m∠BAR = 81°
