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,

\(\Rightarrow RP^2 \ = \ RQ^2 \ + \ QP^2 \)

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

\( \Rightarrow 625\ - \ 50x \ + \ x^2 \ = \ x^2 \ + \ 25 \)

\( \Rightarrow - 50x \ = \ -600 \)

\( \Rightarrow x \ = \ \frac{600}{50} \ = \ 12 \)

\(\Rightarrow \) RQ = 12 cm

\(\Rightarrow \) 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 \( tanP \ = \ \frac{P}{B} \ = \ \frac{RQ}{PQ} \ = \ \frac{12}{5} \)