Solution:
Given the triangle
Let h represent the hypotenuse
[tex]\begin{gathered} h^2=8^2+15^2\text{ \lparen pythagoras theorem\rparen} \\ h^2=64+225 \\ h^2=289 \\ h=\sqrt{289} \\ h=17 \end{gathered}[/tex]
[tex](a)\text{ }sin\theta=\frac{opposite}{hypotenuse\text{ }}\text{ = }\frac{15}{17}[/tex][tex](b)\text{ cos}\theta=\frac{adjacent}{hypotenuse}=\frac{8}{17}[/tex][tex](c)\text{ Tan}\theta=\frac{opposite\text{ }}{adjacent}=\frac{15}{8}[/tex][tex](d)\text{ Csc}\theta=\frac{1}{sin\theta}=\frac{1}{\frac{15}{17}}\text{ = }\frac{17}{15}[/tex][tex](e)\text{ Sec}\theta=\frac{1}{cos\theta}=\frac{1}{\frac{8}{17}}=\frac{17}{8}[/tex][tex](f)\text{ Cot}\theta=\frac{1}{tan\theta}=\frac{1}{\frac{15}{8}}=\frac{8}{15}[/tex]
