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
