To find the sine ratio of the triangle given that \(\cos(\theta) = \frac{\sqrt{2}}{5}\), you can use the Pythagorean identity:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
First, substitute the value of \(\cos(\theta)\):
\[ \sin^2(\theta) + \left(\frac{\sqrt{2}}{5}\right)^2 = 1 \]
Calculating \(\left(\frac{\sqrt{2}}{5}\right)^2\):
\[ \sin^2(\theta) + \frac{2}{25} = 1 \]
Next, isolate \(\sin^2(\theta)\):
\[ \sin^2(\theta) = 1 - \frac{2}{25} \]
Convert 1 into a fraction with the same denominator:
\[ 1 = \frac{25}{25} \]
Now, perform the subtraction:
\[ \sin^2(\theta) = \frac{25}{25} - \frac{2}{25} = \frac{23}{25} \]
Now take the square root to find \(\sin(\theta)\):
\[ \sin(\theta) = \pm \sqrt{\frac{23}{25}} = \pm \frac{\sqrt{23}}{5} \]
The sine value will depend on the quadrant in which the angle \(\theta\) is located. If \(\theta\) is in the first quadrant, \(\sin(\theta)\) is positive:
\[ \sin(\theta) = \frac{\sqrt{23}}{5} \]
If \(\theta\) is in the second quadrant, \(\sin(\theta)\) remains positive, so:
\[ \sin(\theta) = \frac{\sqrt{23}}{5} \]
Thus, the sine ratio of the triangle is:
\[ \sin(\theta) = \frac{\sqrt{23}}{5} \]