# In $\triangle P Q R$, right angled at Q,

Question.

In $\triangle P Q R$, right angled at $Q, P R+Q R=25 \mathrm{~cm}$ and $P Q=5 \mathrm{~cm}$. Determine the values of $\sin P, \cos P$ and $\tan P$.

Solution:

In figure,

PQ = 5 cm

PR + QR = 25 cm

i.e., PR = 25 cm – QR

Now, $\mathrm{PR}^{2}=\mathrm{PQ}^{2}+\mathrm{QR}^{2}$

$\Rightarrow \quad(25-\mathrm{QR})^{2}=(5)^{2}+\mathrm{QR}^{2}$

$\Rightarrow \quad 625-50 \times \mathrm{QR}+\mathrm{QR}^{2}=25+\mathrm{QR}^{2}$

50 × QR = 600  QR = 12 cm

and PR = 25 cm – 12cm = 13 cm

We find $\sin P=\frac{Q R}{P R}=\frac{12}{13}, \cos P=\frac{P Q}{P R}=\frac{5}{13}$

and $\tan P=\frac{Q R}{P Q}=\frac{12}{5}$