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If $sin\ A = \frac{3}{4}$, calculate $cos\ A$ and $tan\ A$.
Given:
$sin\ A = \frac{3}{4}$.
To do:
We have to find $cos\ A$ and $tan\ A$.
Solution:
Let, in a triangle $ABC$ right-angled at $B$, $sin\ A = \frac{3}{4}$.
We know that,
In a right-angled triangle $ABC$ with a 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,
$AC^2=AB^2+BC^2$
$\Rightarrow (4)^2=(AB)^2+(3)^2$
$\Rightarrow 16=(AB)^2+9$
$\Rightarrow AB=\sqrt{16-9}=\sqrt7$
Therefore,
$cos\ A=\frac{AB}{AC}$
$=\frac{\sqrt7}{4}$
$tan\ A=\frac{BC}{AB}$
$=\frac{3}{\sqrt7}$
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