Given:
The figure of a right triangle.
To find:
All the six trigonometric ratios for the angle θ.
Solution:
Using the Pythagoras theorem, we get
[tex]Hypotenuse^2=Perpendicular^2+Base^2[/tex]
[tex]Hypotenuse^2=(9)^2+(12)^2[/tex]
[tex]Hypotenuse^2=81+144[/tex]
[tex]Hypotenuse^2=225[/tex]
Taking square root on both sides, we get
[tex]Hypotenuse=15[/tex]
Now, the six trigonometric ratios for the angle θ are:
[tex]\sin\theta=\dfrac{Perpendicular}{Hypotenuse}[/tex]
[tex]\sin\theta=\dfrac{12}{15}[/tex]
[tex]\sin\theta=\dfrac{4}{5}[/tex]
[tex]\cos\theta=\dfrac{Base}{Hypotenuse}[/tex]
[tex]\cos\theta=\dfrac{9}{15}[/tex]
[tex]\cos\theta=\dfrac{3}{5}[/tex]
[tex]\tan\theta=\dfrac{Perpendicular}{base}[/tex]
[tex]\tan\theta=\dfrac{12}{9}[/tex]
[tex]\tan\theta=\dfrac{4}{3}[/tex]
[tex]\csc\theta=\dfrac{1}{\sin \theta}[/tex]
[tex]\csc\theta=\dfrac{1}{\dfrac{4}{5}}[/tex]
[tex]\csc\theta=\dfrac{5}{4}[/tex]
[tex]\sec\theta=\dfrac{1}{\cos \theta}[/tex]
[tex]\sec\theta=\dfrac{1}{\dfrac{3}{5}}[/tex]
[tex]\sec\theta=\dfrac{5}{3}[/tex]
[tex]\cot\theta=\dfrac{1}{\tan \theta}[/tex]
[tex]\cot\theta=\dfrac{1}{\dfrac{4}{3}}[/tex]
[tex]\cot\theta=\dfrac{3}{4}[/tex]
Therefore, the six trigonometric ratios are [tex]\sin\theta=\dfrac{4}{5},\cos\theta=\dfrac{3}{5},\tan\theta=\dfrac{4}{3},\csc\theta=\dfrac{5}{4},\sec\theta=\dfrac{5}{3},\cot\theta=\dfrac{3}{4}[/tex].
