Mark y inside regular pentagon pqrst so that pqy is equilateral is ryt straight? explain

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.