Maria correctly concluded that not enough information was provided to construct the requested proof. What extra information would allow Maria to contrast the proof? SA=TR. RS= ST. ASCT is similar to ASCA. SC=5/2BR

Hello your question is incomplete attached below is the complete question

RS = ST ( option b )

Step-by-step explanation:

First we consider Δ SCA and Δ TCB

∠ SCA  = ∠ TCB  ,  ∠CAS = ∠CBT = 90°

if we assume  RS = ST then ΔSRT will be termed an Isosceles triangle with its legs : SR = ST. also vertex S bisects the base of the triangle.

Δ SBR ≅ ΔSBT  from here we can deduce that ; BR = BT

From the assumptions what is actually missing to contrast the proof is : RS = ST

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47

explanation:

step-by-step explanation: