Answer:
Part 1) ![sin(\theta)=-\frac{\sqrt{23}}{5}](https://tex.z-dn.net/?f=sin%28%5Ctheta%29%3D-%5Cfrac%7B%5Csqrt%7B23%7D%7D%7B5%7D)
Part 2) ![tan(\theta)=\frac{\sqrt{46}}{2}](https://tex.z-dn.net/?f=tan%28%5Ctheta%29%3D%5Cfrac%7B%5Csqrt%7B46%7D%7D%7B2%7D)
Step-by-step explanation:
step 1
Find the ![sin(\theta)](https://tex.z-dn.net/?f=sin%28%5Ctheta%29)
we know that
![sin^{2}(\theta) +cos^{2}(\theta)=1](https://tex.z-dn.net/?f=sin%5E%7B2%7D%28%5Ctheta%29%20%2Bcos%5E%7B2%7D%28%5Ctheta%29%3D1)
we have
![cos(\theta)=-\frac{\sqrt{2}}{5}](https://tex.z-dn.net/?f=cos%28%5Ctheta%29%3D-%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B5%7D)
substitute
![sin^{2}(\theta) +(-\frac{\sqrt{2}}{5})^{2}=1](https://tex.z-dn.net/?f=sin%5E%7B2%7D%28%5Ctheta%29%20%2B%28-%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B5%7D%29%5E%7B2%7D%3D1)
![sin^{2}(\theta) +\frac{2}{25}=1](https://tex.z-dn.net/?f=sin%5E%7B2%7D%28%5Ctheta%29%20%2B%5Cfrac%7B2%7D%7B25%7D%3D1)
![sin^{2}(\theta)=1-\frac{2}{25}](https://tex.z-dn.net/?f=sin%5E%7B2%7D%28%5Ctheta%29%3D1-%5Cfrac%7B2%7D%7B25%7D)
![sin^{2}(\theta)=\frac{23}{25}](https://tex.z-dn.net/?f=sin%5E%7B2%7D%28%5Ctheta%29%3D%5Cfrac%7B23%7D%7B25%7D)
square root both sides
![sin(\theta)=\pm\frac{\sqrt{23}}{5}](https://tex.z-dn.net/?f=sin%28%5Ctheta%29%3D%5Cpm%5Cfrac%7B%5Csqrt%7B23%7D%7D%7B5%7D)
Remember that the angle θ terminates in Quadrant III
That means, that the value of sin(θ) is negative
so
![sin(\theta)=-\frac{\sqrt{23}}{5}](https://tex.z-dn.net/?f=sin%28%5Ctheta%29%3D-%5Cfrac%7B%5Csqrt%7B23%7D%7D%7B5%7D)
step 2
Find the ![tan(\theta)](https://tex.z-dn.net/?f=tan%28%5Ctheta%29)
we know that
![tan(\theta)=\frac{sin(\theta)}{cos(\theta)}](https://tex.z-dn.net/?f=tan%28%5Ctheta%29%3D%5Cfrac%7Bsin%28%5Ctheta%29%7D%7Bcos%28%5Ctheta%29%7D)
we have
![sin(\theta)=-\frac{\sqrt{23}}{5}](https://tex.z-dn.net/?f=sin%28%5Ctheta%29%3D-%5Cfrac%7B%5Csqrt%7B23%7D%7D%7B5%7D)
![cos(\theta)=-\frac{\sqrt{2}}{5}](https://tex.z-dn.net/?f=cos%28%5Ctheta%29%3D-%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B5%7D)
substitute
![tan(\theta)=-\frac{\sqrt{23}}{5}:-\frac{\sqrt{2}}{5}=\frac{\sqrt{23}}{\sqrt{2}}](https://tex.z-dn.net/?f=tan%28%5Ctheta%29%3D-%5Cfrac%7B%5Csqrt%7B23%7D%7D%7B5%7D%3A-%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B5%7D%3D%5Cfrac%7B%5Csqrt%7B23%7D%7D%7B%5Csqrt%7B2%7D%7D)
simplify
![tan(\theta)=\frac{\sqrt{46}}{2}](https://tex.z-dn.net/?f=tan%28%5Ctheta%29%3D%5Cfrac%7B%5Csqrt%7B46%7D%7D%7B2%7D)
115 days since august 24th 2017
Π=3.14159265359 so I guess 1, is the answer
Answer:
∠L=23°
Step-by-step explanation:
(6x-5)+(x+9)=(9x-24) (ext angle of triangle)
6x-5+x+9=9x-24
7x+4=9x-24
7x-9x=-24-4
-2x=-28
2x=28
x=28÷2
=14
Hence, ∠L = (x+9)°
= (14+9)°
= 23°
Answer: B.
![-8xy + 12x + y^2 - 2y](https://tex.z-dn.net/?f=-8xy%20%2B%2012x%20%2B%20y%5E2%20-%202y)
Step-by-step explanation:
i just took the assignment