But its still not that much info what is the answers for it if there is no answer
How are you finding what does x equal
Answer:
![(a)\ \sec^2(\theta) = 82](https://tex.z-dn.net/?f=%28a%29%5C%20%5Csec%5E2%28%5Ctheta%29%20%3D%2082)
![(b)\ \cot(\theta) = \frac{1}{9}](https://tex.z-dn.net/?f=%28b%29%5C%20%5Ccot%28%5Ctheta%29%20%3D%20%5Cfrac%7B1%7D%7B9%7D)
![(c)\ \cot(\frac{\pi}{2} - \theta) = 9](https://tex.z-dn.net/?f=%28c%29%5C%20%5Ccot%28%5Cfrac%7B%5Cpi%7D%7B2%7D%20-%20%5Ctheta%29%20%3D%209)
![(d)\ \csc^2(\theta) = \frac{82}{81}](https://tex.z-dn.net/?f=%28d%29%5C%20%5Ccsc%5E2%28%5Ctheta%29%20%3D%20%5Cfrac%7B82%7D%7B81%7D)
Step-by-step explanation:
Given
![\tan(\theta) = 9](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%29%20%3D%209)
Required
Solve (a) to (d)
Using tan formula, we have:
![\tan(\theta) = \frac{Opposite}{Adjacent}](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%29%20%3D%20%5Cfrac%7BOpposite%7D%7BAdjacent%7D)
This gives:
![\frac{Opposite}{Adjacent} = 9](https://tex.z-dn.net/?f=%5Cfrac%7BOpposite%7D%7BAdjacent%7D%20%3D%209)
Rewrite as:
![\frac{Opposite}{Adjacent} = \frac{9}{1}](https://tex.z-dn.net/?f=%5Cfrac%7BOpposite%7D%7BAdjacent%7D%20%3D%20%5Cfrac%7B9%7D%7B1%7D)
Using a unit ratio;
![Opposite = 9; Adjacent = 1](https://tex.z-dn.net/?f=Opposite%20%3D%209%3B%20Adjacent%20%3D%201)
Using Pythagoras theorem, we have:
![Hypotenuse^2 = Opposite^2 + Adjacent^2](https://tex.z-dn.net/?f=Hypotenuse%5E2%20%3D%20Opposite%5E2%20%2B%20Adjacent%5E2)
![Hypotenuse^2 = 9^2 + 1^2](https://tex.z-dn.net/?f=Hypotenuse%5E2%20%3D%209%5E2%20%2B%201%5E2)
![Hypotenuse^2 = 81 + 1](https://tex.z-dn.net/?f=Hypotenuse%5E2%20%3D%2081%20%2B%201)
![Hypotenuse^2 = 82](https://tex.z-dn.net/?f=Hypotenuse%5E2%20%3D%2082)
Take square roots of both sides
![Hypotenuse =\sqrt{82}](https://tex.z-dn.net/?f=Hypotenuse%20%3D%5Csqrt%7B82%7D)
So, we have:
![Opposite = 9; Adjacent = 1](https://tex.z-dn.net/?f=Opposite%20%3D%209%3B%20Adjacent%20%3D%201)
![Hypotenuse =\sqrt{82}](https://tex.z-dn.net/?f=Hypotenuse%20%3D%5Csqrt%7B82%7D)
Solving (a):
![\sec^2(\theta)](https://tex.z-dn.net/?f=%5Csec%5E2%28%5Ctheta%29)
This is calculated as:
![\sec^2(\theta) = (\sec(\theta))^2](https://tex.z-dn.net/?f=%5Csec%5E2%28%5Ctheta%29%20%3D%20%28%5Csec%28%5Ctheta%29%29%5E2)
![\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2](https://tex.z-dn.net/?f=%5Csec%5E2%28%5Ctheta%29%20%3D%20%28%5Cfrac%7B1%7D%7B%5Ccos%28%5Ctheta%29%7D%29%5E2)
Where:
![\cos(\theta) = \frac{Adjacent}{Hypotenuse}](https://tex.z-dn.net/?f=%5Ccos%28%5Ctheta%29%20%3D%20%5Cfrac%7BAdjacent%7D%7BHypotenuse%7D)
![\cos(\theta) = \frac{1}{\sqrt{82}}](https://tex.z-dn.net/?f=%5Ccos%28%5Ctheta%29%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7B82%7D%7D)
So:
![\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2](https://tex.z-dn.net/?f=%5Csec%5E2%28%5Ctheta%29%20%3D%20%28%5Cfrac%7B1%7D%7B%5Ccos%28%5Ctheta%29%7D%29%5E2)
![\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2](https://tex.z-dn.net/?f=%5Csec%5E2%28%5Ctheta%29%20%3D%20%28%5Cfrac%7B1%7D%7B%5Cfrac%7B1%7D%7B%5Csqrt%7B82%7D%7D%7D%29%5E2)
![\sec^2(\theta) = (\sqrt{82})^2](https://tex.z-dn.net/?f=%5Csec%5E2%28%5Ctheta%29%20%3D%20%28%5Csqrt%7B82%7D%29%5E2)
![\sec^2(\theta) = 82](https://tex.z-dn.net/?f=%5Csec%5E2%28%5Ctheta%29%20%3D%2082)
Solving (b):
![\cot(\theta)](https://tex.z-dn.net/?f=%5Ccot%28%5Ctheta%29)
This is calculated as:
![\cot(\theta) = \frac{1}{\tan(\theta)}](https://tex.z-dn.net/?f=%5Ccot%28%5Ctheta%29%20%3D%20%5Cfrac%7B1%7D%7B%5Ctan%28%5Ctheta%29%7D)
Where:
---- given
So:
![\cot(\theta) = \frac{1}{\tan(\theta)}](https://tex.z-dn.net/?f=%5Ccot%28%5Ctheta%29%20%3D%20%5Cfrac%7B1%7D%7B%5Ctan%28%5Ctheta%29%7D)
![\cot(\theta) = \frac{1}{9}](https://tex.z-dn.net/?f=%5Ccot%28%5Ctheta%29%20%3D%20%5Cfrac%7B1%7D%7B9%7D)
Solving (c):
![\cot(\frac{\pi}{2} - \theta)](https://tex.z-dn.net/?f=%5Ccot%28%5Cfrac%7B%5Cpi%7D%7B2%7D%20-%20%5Ctheta%29)
In trigonometry:
![\cot(\frac{\pi}{2} - \theta) = \tan(\theta)](https://tex.z-dn.net/?f=%5Ccot%28%5Cfrac%7B%5Cpi%7D%7B2%7D%20-%20%5Ctheta%29%20%3D%20%5Ctan%28%5Ctheta%29)
Hence:
![\cot(\frac{\pi}{2} - \theta) = 9](https://tex.z-dn.net/?f=%5Ccot%28%5Cfrac%7B%5Cpi%7D%7B2%7D%20-%20%5Ctheta%29%20%3D%209)
Solving (d):
![\csc^2(\theta)](https://tex.z-dn.net/?f=%5Ccsc%5E2%28%5Ctheta%29)
This is calculated as:
![\csc^2(\theta) = (\csc(\theta))^2](https://tex.z-dn.net/?f=%5Ccsc%5E2%28%5Ctheta%29%20%3D%20%28%5Ccsc%28%5Ctheta%29%29%5E2)
![\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2](https://tex.z-dn.net/?f=%5Ccsc%5E2%28%5Ctheta%29%20%3D%20%28%5Cfrac%7B1%7D%7B%5Csin%28%5Ctheta%29%7D%29%5E2)
Where:
![\sin(\theta) = \frac{Opposite}{Hypotenuse}](https://tex.z-dn.net/?f=%5Csin%28%5Ctheta%29%20%3D%20%5Cfrac%7BOpposite%7D%7BHypotenuse%7D)
![\sin(\theta) = \frac{9}{\sqrt{82}}](https://tex.z-dn.net/?f=%5Csin%28%5Ctheta%29%20%3D%20%5Cfrac%7B9%7D%7B%5Csqrt%7B82%7D%7D)
So:
![\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2](https://tex.z-dn.net/?f=%5Ccsc%5E2%28%5Ctheta%29%20%3D%20%28%5Cfrac%7B1%7D%7B%5Cfrac%7B9%7D%7B%5Csqrt%7B82%7D%7D%7D%29%5E2)
![\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2](https://tex.z-dn.net/?f=%5Ccsc%5E2%28%5Ctheta%29%20%3D%20%28%5Cfrac%7B%5Csqrt%7B82%7D%7D%7B9%7D%29%5E2)
![\csc^2(\theta) = \frac{82}{81}](https://tex.z-dn.net/?f=%5Ccsc%5E2%28%5Ctheta%29%20%3D%20%5Cfrac%7B82%7D%7B81%7D)
70: 1,2,5,7,10,14,35,<span>70
20: 1,2,4,5,10,20</span>
Answer:
200 = 10 Hours, 400 = 5 Hours, 600 = 3 Hours & 18 Minutes.
Step-by-step explanation:
2000 / How many miles it travels per hour.
2000 / 200 = 10
2000 / 400 = 5
2000 / 600 = 3.3..