10. In ∆ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P

In a given triangle PQR, right angled at Q, the following measures are PQ = 5 cm PR + QR = 25 cm Now let us assume, QR = x PR = 25-QR PR = 25- x According to the Pythagorean Theorem, PR2 = PQ2 + QR2 Substitute the value of PR as x (25- x) 2 = 52 + x2 252 + x2 – 50x = 25 + x2 625 + x2-50x -25 – x2 = 0 -50x = -600 x= -600/-50 x = 12 = QR Now, find the value of PR PR = 25- QR Substitute the value of QR PR = 25-12 PR = 13 Now, substitute the value to the given problem (1) sin p = Opposite Side/Hypotenuse = QR/PR = 12/13 (2) Cos p = Adjacent Side/Hypotenuse = PQ/PR = 5/13 (3) tan p =Opposite Side/Adjacent side = QR/PQ = 12/5