In ∆PQR, right angled at Q, PQ = 4 cm and RQ = 3 cm.

Given:

$\triangle P Q R$ is right angled at vertex $Q$.

$P Q=4 \mathrm{~cm}$

$R Q=3 \mathrm{~cm}$

To find:

$\sin P, \sin R, \sec P, \sec R$

Given $\triangle P Q R$ is as shown below

Hypotenuse side PR is unknown.

Therefore, we find side $P R$ of $\triangle P Q R$ by Pythagoras theorem

By applying Pythagoras theorem to $\triangle P Q R$

We get,

$P R^{2}=P Q^{2}+R Q^{2}$

Substituting values of sides from the above figure

$P R^{2}=4^{2}+3^{2}$

$P R^{2}=16+9$

$P R^{2}=25$

$P R=\sqrt{25}$

$P R=5$

Hence, Hypotenuse = 5

Now by definition,

$\sin P=\frac{\text { Perpendicular side opposite to } \angle P}{\text { Hypotenuse }}$

$\sin P=\frac{R Q}{P R}$

Substituting values of sides from the above figure

$\sin P=\frac{3}{5}$

Now by definition,

$\sin R=\frac{\text { Perpendicular side opposite to } \angle R}{\text { Hypotenuse }}$

$\sin R=\frac{P Q}{P R}$

Substituting values of sides from the above figure

$\sin R=\frac{4}{5}$

By definition,

$\sec P=\frac{1}{\cos P}$

$\sec P=\frac{1}{\frac{\text { Base side adjacent to } \angle P}{\text { Hypotenuse }}}$

$\sec P=\frac{\text { Hypotenuse }}{\text { Base side adjacent to } \angle P} .$

Substituting values of sides from the above figure

$\sec P=\frac{\mathrm{PR}}{P Q}$

By definition,

$\sec R=\frac{1}{\cos R}$

$\sec R=\frac{1}{\frac{\text { Base side adjacent to } \angle R}{\text { Hypotenuse }}}$

$\sec R=\frac{\text { Hypotenuse }}{\text { Base side adjacent to } \angle R} .$

Substituting values of sides from the above figure

$\sec R=\frac{\mathrm{PR}}{R Q}$

$\sec R=\frac{5}{3}$

Answer: $\sin P=\frac{3}{5}, \sin R=\frac{4}{5}, \sec P=\frac{5}{4}$ and $\sec R=\frac{5}{3}$