Answer:
a)
b) The other 5 trigonometric ratios are:
[tex]\begin{gathered} \sin\theta=\frac{3\sqrt{58}}{58} \\ \\ \\ \\ \end{gathered}[/tex][tex]\cos\theta=\frac{7\sqrt{58}}{58}[/tex][tex]\csc\theta=\frac{\sqrt{58}}{3}[/tex][tex]\sec\theta=\frac{\sqrt{58}}{7}[/tex][tex]\ctg\theta=\frac{7}{3}[/tex]
Explanation:
The in a right triangle of the form:
the tan ratio is:
[tex]\tan\theta=\frac{opposite}{adjacent}[/tex]
Then, we know:
[tex]\tan\theta=\frac{3}{7}[/tex]
Then,
opposite = 7
adjacent = 3
We can draw:
For b, we need to know the formulas for the other 5 trigonometric ratios.
The formulas are:
[tex]\begin{gathered} \sin(\theta)=\frac{opposite}{hypotenuse} \\ . \\ \cos(\theta)=\frac{adjacent}{hypotenuse} \\ . \\ \csc(\theta)=\frac{1}{\sin(\theta)} \\ . \\ \sec(\theta)=\frac{1}{\cos(\theta)} \\ . \\ \ctg(\theta)=\frac{1}{\tan(\theta)} \end{gathered}[/tex]
We can see that we need to know the length of the hypotenuse. We can use the pythagoren theorem to find it.
The pytagorean theorem tell us:
[tex]Hypotenuse^2=opposite^2+adjacent^2[/tex]
Then:
[tex]Hypotenuse^2=3^2+7^2[/tex]
And solve:
[tex]\begin{gathered} Hypotenuse^2=9+49=58 \\ Hypotenuse=\sqrt{58} \end{gathered}[/tex]
Now we can find:
[tex]\sin(\theta)=\frac{oppos\imaginaryI te}{hypotenuse}=\frac{3}{\sqrt{58}}[/tex]
We need to radicalize the denominator. Thus:
