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