Find the reference angle given: t = −76°

This angle t= -76º lies on the IV quadrant, so the reference angle is the acute positive angle fron this angle to x-axis, so in this case the reference angle is 76º. (if better if you can draw it).
Because, the formula for this quadrant is: Ref angle= 360º - given angle (but given angle has to be positive so for -76º turned positive is 284º) now you can apply the formula Ref angle= 360º - 284º= 76º

2.4+3.2t. you get the answer by adding the ones with variables and ones without