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,triangle

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 = \(QR\over PQ\) = \(12\over 5\)

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