# In $$\triangle$$ PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

## 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 = $$QR\over PQ$$ = $$12\over 5$$