If $sin\ A = \frac{9}{41}$, compute $cos\ A$ and $tan\ A$.

Given:

$sin\ A = \frac{9}{41}$.

To do:

We have to compute $cos\ A$ and $tan\ A$.

Solution:  

Let, in a triangle $ABC$ right angled at $B$, $sin\ A = \frac{9}{41}$.

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,

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

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

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

Here,

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

$\Rightarrow (41)^2=AB^2+(9)^2$

$\Rightarrow AB^2=1681-81$

$\Rightarrow AB=\sqrt{1600}=40$

Therefore,

$cos\ A=\frac{AB}{AC}=\frac{40}{41}$

 $tan\ A=\frac{BC}{AB}=\frac{9}{40}$