Given:
The figure of a right triangle ABC.
To find:
The trigonometric ratios [tex]\sin B, \tan A, \cos B[/tex].
Solution:
Using Pythagoras theorem,
[tex]Hypotenuse^2=Perpendicular^2+Base^2[/tex]
[tex]AB^2=AC^2+BC^2[/tex]
[tex](10)^2=(6)^2+BC^2[/tex]
[tex]100-36=BC^2[/tex]
[tex]64=BC^2[/tex]
Taking square root on both sides.
[tex]\sqrt{64}=BC[/tex]
[tex]8=BC[/tex]
So, measure of BC is 8 units.
Now,
[tex]\sin \theta=\dfrac{Opposite}{Hypotenuse}[/tex]
[tex]\sin B=\dfrac{AC}{AB}[/tex]
[tex]\sin B=\dfrac{6}{10}[/tex]
[tex]\sin B=\dfrac{3}{5}[/tex]
Similarly,
[tex]\tan \theta=\dfrac{Opposite}{Adjacent}[/tex]
[tex]\tan A=\dfrac{AC}{BC}[/tex]
[tex]\tan A=\dfrac{6}{8}[/tex]
[tex]\tan A=\dfrac{3}{4}[/tex]
And,
[tex]\cos \theta=\dfrac{Adjacent}{Hypotenuse}[/tex]
[tex]\cos B=\dfrac{BC}{AB}[/tex]
[tex]\cos B=\dfrac{8}{10}[/tex]
[tex]\cos B=\dfrac{4}{5}[/tex]
Therefore, the required trigonometric ration are [tex]\sin B=\dfrac{3}{5}, \tan A=\dfrac{3}{4}, \cos B=\dfrac{4}{5}[/tex].