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