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

Given:

$15\ cot\ A = 8$.

To do:

We have to find $sin\ A$ and $sec\ A$.

Solution:  

Let, in a triangle $ABC$ right-angled at $B$, $15\ cot\ A = 8$.

This implies,

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

We know that,

In a right-angled triangle $ABC$ with right angle at $B$,

By Pythagoras theorem,

$AC^2=AB^2+BC^2$

By trigonometric ratios definitions,

$sin\ A=\frac{Opposite}{Hypotenuse}=\frac{BC}{AC}$

$cot\ A=\frac{Adjacent}{Opposite}=\frac{AB}{BC}$

$sec\ A=\frac{Hypotenuse}{Adjacent}=\frac{AC}{AB}$

Here,

$AC^2=AB^2+BC^2$

$\Rightarrow AC^2=(8)^2+(15)^2$

$\Rightarrow AC^2=64+225$

$\Rightarrow AC=\sqrt{289}=17$

Therefore,

$sin\ A=\frac{BC}{AC}=\frac{15}{17}$

$sec\ A=\frac{AC}{AB}=\frac{17}{8}$

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Updated on: 10-Oct-2022

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