Question : In a right-angled triangle for an acute angle $x$, find $\sin x$. It is given that $\tan x = 5$.
Option 1: $\frac{1}{3}$
Option 2: $\frac{5}{\sqrt{26}}$
Option 3: $\frac{2}{5}$
Option 4: $\frac{5}{\sqrt{28}}$
Correct Answer: $\frac{5}{\sqrt{26}}$
Solution :
Given,
$\tan x = 5$
We know that, $\tan x=\frac{\text{Perpendicular}}{\text{Base}}=\frac{5}{1}$
Let the perpendicular($p$) and base($b$) be 5 units and 1 unit.
Let the hypotenuse be $h$.
Using the Pythagoras theorem,
$h^2=p^2+b^2$
⇒ $h^2=5^2+1^2$
⇒ $h=\sqrt{26}$ units
So, $\sin x =\frac{\text{Perpendicular}}{\text{Hypotenuse}}= \frac{5}{\sqrt{26}}$
Hence, the correct answer is $\frac{5}{\sqrt{26}}$.
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