Given $15 \cot A = 8$,find $\sin A$ and $\sec A$.

(N/A) Consider a right-angled triangle $ABC$,right-angled at $B$.
$\cot A = \frac{\text{Side adjacent to } \angle A}{\text{Side opposite to } \angle A} = \frac{AB}{BC}$
It is given that $15 \cot A = 8$,so $\cot A = \frac{8}{15}$.
Therefore,$\frac{AB}{BC} = \frac{8}{15}$.
Let $AB = 8k$ and $BC = 15k$,where $k$ is a positive constant.
Applying the Pythagoras theorem in $\triangle ABC$:
$AC^2 = AB^2 + BC^2$
$AC^2 = (8k)^2 + (15k)^2$
$AC^2 = 64k^2 + 225k^2 = 289k^2$
$AC = 17k$
Now,$\sin A = \frac{\text{Side opposite to } \angle A}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{15k}{17k} = \frac{15}{17}$.
And,$\sec A = \frac{\text{Hypotenuse}}{\text{Side adjacent to } \angle A} = \frac{AC}{AB} = \frac{17k}{8k} = \frac{17}{8}$.