Solution :
We have,
PQ = 5 cm
PR + QR = 25 cm ………..(1)
In triangle PQR, By Pythagoras Theorem,
\({PR}^2\) = \({PQ}^2\) + \({QR}^2\)
\(\implies\) \({PQ}^2\) = \({PR}^2\) – \({QR}^2\)
\(\implies\) \({PQ}^2\) = (PR + QR)(PR – QR)
\(\implies\) \(5^2\) = (PR – QR). 25
\(\implies\) PR – QR = 1 ………..(2)
Solving (1) and (2), we have
PR = 13 cm and QR = 12 cm
sin P = \(QR\over PR\) = \(12\over 13\)
cos P = \(PQ\over PR\) = \(5\over 13\)
tan P = \(QR\over PQ\) = \(12\over 5\)
