m∠PQR=82º
1) According to the properties of a Rhombus, the line segment QS bisects the angle ∠PQR into two congruent angles ∠PQT and ∠RQT
So we can write
∠PQT = ∠RQT
Let's find the value of x:
4x-27= 2x +7
4x -2x = 7 +27
2x = 34
x= 17
2) So m∠PQR is going to be the sum of m∠PQT and m∠RQT
m∠PQR=m∠PQT +m∠RQT
m∠PQR = 4x -27 +2x +7 Plug into that the value for x=17
m∠PQR =4(17) -27 + 2(17) +7
m∠PQR =68-27+34+7
m∠PQR=82º
3) So the answer is m∠PQR=82º