Given 15 cot A = 8, find sin A and sec A.
Given: $15 \cot A=8$
To Find: $\sin A, \sec A$
Since $15 \cot A=8$
By taking 15 on R.H.S
We get,
$\cot A=\frac{8}{15} \ldots \ldots .(1)$
By definition,
$\cot A=\frac{1}{\tan A}$
Hence,
$\cot A=\frac{1}{\frac{\text { Perpendicular side opposite to } \angle \mathrm{A}}{\text { Base side adjacent to } \angle \mathrm{A}}}$
$\cot A=\frac{\text { Base side adjacent to } \angle \mathrm{A}}{\text { Perpendicular side opposite to } \angle \mathrm{A}}$.....(2)
Comparing equation (1) and (2)
We get,
Base side adjacent to $\angle \mathrm{A}=8$
Perpendicular side opposite to $\angle \mathrm{A}=15$
$\triangle A B C$ can be drawn as shown below using above information
Hypotenuse side AC is unknown.
Therefore, we find side $\mathrm{AC}$ of $\triangle A B C$ by Pythagoras theorem.
So, by applying Pythagoras theorem to $\triangle A B C$
We get,
$A C^{2}=A B^{2}+B C^{2}$
Substituting values of sides from the above figure
$A C^{2}=8^{2}+15^{2}$
$A C^{2}=64+225$
$A C^{2}=289$
$A C=\sqrt{289}$
$A C=17$
Therefore, Hypotenuse = 17
Now by definition,
$\sin A=\frac{\text { Perpendicular side opposite to } \angle \mathrm{A}}{\text { Hypotenuse }}$
Therefore, $\sin A=\frac{B C}{A C}$
Substituting values of sides from the above figure
$\sin A=\frac{15}{17}$
By definition,
$\sec A=\frac{1}{\cos A}$
Hence,
$\sec A=\frac{1}{\frac{\text { Base side adjacent to } \angle \mathrm{A}}{\text { Hypotenuse }}}$
$\sec A=\frac{\text { Hypotenuse }}{\text { Base side adjacent to } \angle \mathrm{A}}$.
Substituting values of sides from the above figure
$\sec A=\frac{17}{8}$
Answer: $\sin A=\frac{15}{17}$ and $\sec A=\frac{17}{8}$
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