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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?

Mar 30, 2018

#1
+100595
+3

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

Mar 30, 2018
#2
+7601
+4

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°

Mar 30, 2018
#3
+100595
+2

I like your way better, hectictar  !!!!

CPhill  Mar 30, 2018
#4
+4221
+2

Thanks so much, guys! Bravo!

Mar 31, 2018