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)
J = 106% . x
S = 106% . (550+x)
O = 106% . 2x
S = 2(212+O) = 106%(550+x)
S = 424 + 2O = 583 + 106%x
2(106%.2x) - 106%x = 583-424
159 = 106%x( 4 - 1 )
159 = 106%x ( 3 )
159 = 318%x
1 = 318%x : 159
1 = 2%x
1 = 2/100 . x
x = 50
Jivesh put $50
Stacy put $600
Oliver put $100
Answer:
Step-by-step explanation:
Use trigonometry
<u>The ratio of the adjacent side of a right triangle to the hypotenuse is called the cosine:</u>
- cos x = 18/25 = 0.72
- cos x = 0.72
- x = arccos 0.72
- x = 43.9455° ≈ 44.0° rounded
The value of this equation is 20.
Answer = 16/3
4/3 x 4/1 = 16/3