Since ΔPYQ is equilateral, all 3 angles are 60°. In a regular pentagon, there are (5-2)(180)=3(180)=540° total interior. Divide this by 5 and we get 108° for each interior angle. This means that ∠TPY = 108-60 = 48°. Since PY = PT, this makes ΔPYT isosceles. It also means that ∠PTY = ∠PYT. There are 180° in a triangle; taking out the measure of ∠TPY we have 180-48=132 for these two angles. Divide by 2 and we know that ∠PTY = ∠PYT = 66°.
By the same logic, ∠Q in our pentagon is 108°. Since ΔPYQ is equilateral, ∠PQY = 60°. That means that ∠YQR = 108-60 = 48°. Since QR = QY, ΔQYR is isosceles, which means the base angles are congruent. We take 180-48 = 132 and divide that by 2 to get each base angle, ∠QYR and ∠QRY, is 66°. Looking at the 3 angles around point Y: If ∠RYT is a straight angle, these 3 along this path should add up to 180°. We have ∠PYT=66° + ∠PYQ=60° + ∠QYR=66° for a total of 192°. Therefore ∠RYT is not a straight angle.