1. The given rectangular equation is
.
We substitute
.
![r\cos \theta=2](https://tex.z-dn.net/?f=r%5Ccos%20%5Ctheta%3D2)
Divide through by ![\cos \theta](https://tex.z-dn.net/?f=%5Ccos%20%5Ctheta)
![r=\frac{2}{\cos \theta}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B2%7D%7B%5Ccos%20%5Ctheta%7D)
![r=2}\sec \theta](https://tex.z-dn.net/?f=r%3D2%7D%5Csec%20%5Ctheta)
![\boxed{x=2\to r=2\sec \theta}](https://tex.z-dn.net/?f=%5Cboxed%7Bx%3D2%5Cto%20r%3D2%5Csec%20%5Ctheta%7D)
2. The given rectangular equation is:
![x^2+y^2=36](https://tex.z-dn.net/?f=x%5E2%2By%5E2%3D36)
This is the same as:
![x^2+y^2=6^2](https://tex.z-dn.net/?f=x%5E2%2By%5E2%3D6%5E2)
We use the relation ![r^2=x^2+y^2](https://tex.z-dn.net/?f=r%5E2%3Dx%5E2%2By%5E2)
This implies that:
![r^2=6^2](https://tex.z-dn.net/?f=r%5E2%3D6%5E2)
![\therefore r=6](https://tex.z-dn.net/?f=%5Ctherefore%20r%3D6)
![\boxed{x^2+y^2=36\to r=6}](https://tex.z-dn.net/?f=%5Cboxed%7Bx%5E2%2By%5E2%3D36%5Cto%20r%3D6%7D)
3. The given rectangular equation is:
![x^2+y^2=2y](https://tex.z-dn.net/?f=x%5E2%2By%5E2%3D2y)
This is the same as:
We use the relation
and ![y=r\sin \theta](https://tex.z-dn.net/?f=y%3Dr%5Csin%20%5Ctheta)
This implies that:
![r^2=2r\sin \theta](https://tex.z-dn.net/?f=r%5E2%3D2r%5Csin%20%5Ctheta)
Divide through by r
![r=2\sin \theta](https://tex.z-dn.net/?f=r%3D2%5Csin%20%5Ctheta)
![\boxed{x^2+y^2=2y\to r=2\sin \theta}](https://tex.z-dn.net/?f=%5Cboxed%7Bx%5E2%2By%5E2%3D2y%5Cto%20r%3D2%5Csin%20%5Ctheta%7D)
4. We have ![x=\sqrt{3}y](https://tex.z-dn.net/?f=x%3D%5Csqrt%7B3%7Dy)
We substitute
and ![x=r\cos \theta](https://tex.z-dn.net/?f=x%3Dr%5Ccos%20%5Ctheta)
![r\cos \theta=r\sin \theta\sqrt{3}](https://tex.z-dn.net/?f=r%5Ccos%20%5Ctheta%3Dr%5Csin%20%5Ctheta%5Csqrt%7B3%7D)
This implies that;
![\tan \theta=\frac{\sqrt{3}}{3}](https://tex.z-dn.net/?f=%5Ctan%20%5Ctheta%3D%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B3%7D)
![\theta=\frac{\pi}{6}](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cfrac%7B%5Cpi%7D%7B6%7D)
![\boxed{x=\sqrt{3}y\to \theta=\frac{\pi}{6}}](https://tex.z-dn.net/?f=%5Cboxed%7Bx%3D%5Csqrt%7B3%7Dy%5Cto%20%5Ctheta%3D%5Cfrac%7B%5Cpi%7D%7B6%7D%7D)
5. We have ![x=y](https://tex.z-dn.net/?f=x%3Dy)
We substitute
and ![x=r\cos \theta](https://tex.z-dn.net/?f=x%3Dr%5Ccos%20%5Ctheta)
![r\cos \theta=r\sin \theta](https://tex.z-dn.net/?f=r%5Ccos%20%5Ctheta%3Dr%5Csin%20%5Ctheta)
This implies that;
![\tan \theta=1](https://tex.z-dn.net/?f=%5Ctan%20%5Ctheta%3D1)
![\theta=\frac{\pi}{4}](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cfrac%7B%5Cpi%7D%7B4%7D)
![\boxed{x=y\to \theta=\frac{\pi}{4}}](https://tex.z-dn.net/?f=%5Cboxed%7Bx%3Dy%5Cto%20%5Ctheta%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%7D)
Answer:
2/3 or 4/6
Step-by-step explanation:
Answer:
B.24
Step-by-step explanation:
we know,
exterior angle = 360°/n(where 'n' is number of sides)
then,
15° = 360°/n
n = 360°/15°
n = 24
B. Is the answer to your question
Answer:
39
Step-by-step explanation: