Given:
The terminal point is
[tex](\frac{1}{2},\frac{3}{7})[/tex]
To find:
The six trigonometric ratios.
Explanation:
According to the problem,
The opposite side is
[tex]\frac{3}{7}[/tex]
The adjacent side is
[tex]\frac{1}{2}[/tex]
The hypotenuse side will be,
[tex]\begin{gathered} hyp^2=opp^2+adj^2 \\ hyp^2=(\frac{3}{7})^2+(\frac{1}{2})^2 \\ =\frac{9}{49}+\frac{1}{4} \\ =\frac{36+49}{196} \\ hyp^2=\frac{85}{196} \\ hyp=\sqrt{\frac{85}{196}} \\ hyp=\frac{\sqrt{85}}{\sqrt{196}} \end{gathered}[/tex]
Then, the six trigonometric ratios are,
[tex][/tex]
