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If $ \sin \mathrm{A}=\frac{1}{2} $, then the value of $ \cot \mathrm{A} $ is
(A) $ \sqrt{3} $
(B) $ \frac{1}{\sqrt{3}} $
(C) $ \frac{\sqrt{3}}{2} $
(D) 1
Given:
$sin\ A = \frac{1}{2}$
To do:
We have to find the value of $cot\ A$.
Solution:
Let, in a triangle $ABC$ right-angled at $B$, $sin\ A=\frac{1}{2}$.
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,
$sin\ \theta=\frac{Opposite}{Hypotenuse}=\frac{BC}{AC}$
$cos\ \theta=\frac{Adjacent}{Hypotenuse}=\frac{AB}{AC}$
$sec\ \theta=\frac{Hypotenuse}{Adjacent}=\frac{AC}{AB}$
$tan\ \theta=\frac{Opposite}{Adjacent}=\frac{BC}{AB}$
$cot\ \theta=\frac{Adjacent}{Opposite}=\frac{AB}{BC}$
Here,
$AC^2=AB^2+BC^2$
$\Rightarrow (2)^2=(AB)^2+1^2$
$\Rightarrow AB^2=4-1$
$\Rightarrow AB=\sqrt{3}$
Therefore,
$cot\ \A=\frac{AB}{BC}=\frac{\sqrt3}{1}$
$=\sqrt3$.
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