in right triangle PQR,Q=90DEGREE,PQ=4 AND PR=7 WHAT IS ANGLE R TO THE NEARST TENTH OF A DEGREE

Answer 1

(PQ)^2 + (QR)^2 = (PR)^2,
4^2 + (QR)^2 = 7^2,
(QR)^2 = 7^2 - 4^2 = 49 - 16 = 33,
QR = 5.7.

cosR = QR/PR = 5.7/7 = 0.8207.
R = 34.8 deg.