Q10. 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.

Given, PR + QR = 25 cm and PQ = 5cm

Let QR = x cm
Thus, PR = (25 – x) cm

Then by Pythagoras theorem,

$$RP^2 \ = \ RQ^2 \ + \ QP^2$$

$$(25-x)^2 \ = \ x^2 \ + \ 5^2$$

$$625\ - \ 50x \ + \ x^2 \ = \ x^2 \ + \ 25$$

$$- 50x \ = \ -600$$

=> $$x \ = \ \frac{600}{50} \ = \ 12$$

∴ RQ = 12 cm

=> RP = (25 – 12) = 13 cm

Now, $$sinP \ = \ \frac{P}{H} \ = \ \frac{RQ}{RP} \ = \ \frac{12}{13}$$

$$cosP \ = \ \frac{B}{H} \ = \ \frac{PQ}{RP} \ = \ \frac{5}{13}$$

and $$tanP \ = \ \frac{P}{B} \ = \ \frac{RQ}{PQ} \ = \ \frac{12}{5}$$