Given:
Right triangle
To find:
The six trigonometric functions of θ
Solution:
Hypotenuse = 18
Adjacent side to θ = 10
Opposite side to θ = ?
Using Pythagoras theorem:
![\text {Hypotenuse}^2 = \text{adjacent}^2+\text{opposite}^2](/tpl/images/0579/2489/05249.png)
![18^2 =10^2+\text{opposite}^2](/tpl/images/0579/2489/cb010.png)
![324=100+\text{opposite}^2](/tpl/images/0579/2489/0770f.png)
Subtract 100 from both sides.
![224=\text{opposite}^2](/tpl/images/0579/2489/2965c.png)
Taking square root on both sides.
![4\sqrt{14}=\text{opposite}](/tpl/images/0579/2489/cec6e.png)
Using trigonometric ratio formula:
![$\sin\theta =\frac{\text{opposite }}{\text{hypotenuse}}](/tpl/images/0579/2489/ae4ea.png)
![$\sin\theta =\frac{4\sqrt{14} }{18}](/tpl/images/0579/2489/ef25c.png)
![$\csc\theta =\frac{\text{hypotenuse}}{\text{opposite }}](/tpl/images/0579/2489/90005.png)
![$\csc\theta =\frac{18}{4\sqrt{14} }](/tpl/images/0579/2489/59ee8.png)
![$\cos \theta=\frac{\text { adjacent side }}{\text { hypotenuse }}](/tpl/images/0579/2489/40441.png)
![$\cos \theta=\frac{10}{18}](/tpl/images/0579/2489/a2c45.png)
![$\sec \theta=\frac{\text { hypotenuse }}{\text { adjacent side }}](/tpl/images/0579/2489/1bf89.png)
![$\sec \theta=\frac{18 }{10}](/tpl/images/0579/2489/6e211.png)
![$\tan \theta=\frac{\text { opposite side }}{\text { adjacent side }}](/tpl/images/0579/2489/7ac4e.png)
![$\tan \theta=\frac{4\sqrt{14} }{10}](/tpl/images/0579/2489/8fb2a.png)
![$\cot \theta=\frac{\text { adjacent side }}{\text { opposite side }}](/tpl/images/0579/2489/1b8cc.png)
![$\cot \theta=\frac{10}{4\sqrt{14} }](/tpl/images/0579/2489/19999.png)