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