Question : In a right triangle for an acute angle $x$, if $\sin x=\frac{3}{7}$, then find the value of $\cos x$.
Option 1: $\frac{2}{7}$
Option 2: $\frac{3}{4}$
Option 3: $\frac{1}{\sqrt{3}}$
Option 4: $\frac{2\sqrt{10}}{7}$
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Correct Answer: $\frac{2\sqrt{10}}{7}$
Solution :
Given that, it is a right-angle triangle with an acute angle $x$.
$\therefore$ Angle $x$ lies between $0$ to $\frac{\pi}{2}$
Given: $\sin x=\frac{3}{7}$
Squaring both sides,
$\sin^2 x=\frac{9}{49}$
Since $\sin^2 x+ \cos^2 x=1$,
So, $\cos^2 x=1-\frac{9}{49}$
⇒ $\cos x =\frac{2\sqrt {10}}{7}$
Hence, the correct answer is $\frac{2\sqrt{10}}{7}$.
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