Question.
Given 15 cot A = 8, find sin A and sec A.
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, $A C=\sqrt{(8 k)^{2}+(15 k)^{2}}=17 k$
$\sin A=\frac{B C}{A C}=\frac{15 k}{17 k}=\frac{15}{17}, \quad \sec A=\frac{A C}{A B}=\frac{17 k}{8 k}=\frac{17}{8}$
$\cot A=\frac{8}{15}$
$\Rightarrow \frac{\mathbf{A B}}{\mathbf{B C}}=\frac{\mathbf{8}}{\mathbf{1 5}}$
$\Rightarrow \mathrm{AB}=8 \mathrm{k}$
and $\mathrm{BC}=15 \mathrm{k}$
Now, $A C=\sqrt{(8 k)^{2}+(15 k)^{2}}=17 k$
$\sin A=\frac{B C}{A C}=\frac{15 k}{17 k}=\frac{15}{17}, \quad \sec A=\frac{A C}{A B}=\frac{17 k}{8 k}=\frac{17}{8}$
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