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

Solution:

$\cot A=\frac{8}{15}$

$\Rightarrow \frac{\mathbf{A B}}{\mathbf{B C}}=\frac{\mathbf{8}}{\mathbf{1 5}}$

$\Rightarrow \mathrm{AB}=8 \mathrm{k}$

and $\mathrm{BC}=15 \mathrm{k}$



Now, $A C=\sqrt{(8 k)^{2}+(15 k)^{2}}=17 k$

$\sin A=\frac{B C}{A C}=\frac{15 k}{17 k}=\frac{15}{17}, \quad \sec A=\frac{A C}{A B}=\frac{17 k}{8 k}=\frac{17}{8}$