Answer:
[tex]a.tan(\theta)=\frac{15}{8}[/tex]
[tex]b.csc(\theta)=\frac{17}{15}[/tex]
[tex]d.sec(\theta)=\frac{17}{8}[/tex]
Step-by-step explanation:
The trigonometry functions on a right triangle are given by:
[tex]sin(\theta)=\frac{opposite}{hypotenuse}[/tex]
[tex]csc(\theta)=\frac{hypotenuse}{opposite}[/tex]
[tex]cos(\theta)=\frac{adjacent}{hypotenuse}[/tex]
[tex]sec(\theta)=\frac{hypotenuse}{adjacent}[/tex]
[tex]tan(\theta)=\frac{opposite}{adjacent}[/tex]
[tex]cot(\theta)=\frac{adjacent}{opposite}[/tex]
Let's calculate the adjacent side using pythagorean theorem:
[tex]a^2+b^2=c^2[/tex]
Where:
a=opposite
b=adjacent
c=hypotenuse
The problem provides us a and c because:
[tex]sin(\theta)=\frac{opposite}{hypotenuse}=\frac{15}{17}[/tex]
So:
[tex]15^2+b^2=17^2[/tex]
Solving for b:
[tex]b^2=17^2-15^2\\b^2=289-225\\b^2=64\\b=\sqrt{64} \\b=8[/tex]
Therefore:
a=opposite=15
b=adjacent=8
c=hypotenuse=17
Finally, let's see if the given options are correct:
[tex]a.tan(\theta)=\frac{15}{8}=\frac{opposite}{adjacent}=\frac{15}{8}[/tex] , This is correct.
[tex]b.csc(\theta)=\frac{17}{15}=\frac{adjacent}{hypotenuse}=\frac{17}{15}[/tex] , This is correct.
[tex]c.cot(\theta)=\frac{17}{8}=\frac{adjacent}{opposite}=\frac{8}{15}[/tex] , This is incorrect.
[tex]d.sec(\theta)=\frac{17}{8}=\frac{hypotenuse}{adjacent}=\frac{17}{8}[/tex] , This is correct.
