Answer:
a) ![\cos(\theta) = \frac{\sqrt[]{33}}{7}](https://tex.z-dn.net/?f=%5Ccos%28%5Ctheta%29%20%3D%20%5Cfrac%7B%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D)
b) ![\sin(\theta + \frac{\pi}{6})\frac{-3\sqrt[]{11}+4}{14}](https://tex.z-dn.net/?f=%5Csin%28%5Ctheta%20%2B%20%5Cfrac%7B%5Cpi%7D%7B6%7D%29%5Cfrac%7B-3%5Csqrt%5B%5D%7B11%7D%2B4%7D%7B14%7D)
c) ![\cos(\theta-\pi)=\frac{\sqrt[]{33}}{7}](https://tex.z-dn.net/?f=%5Ccos%28%5Ctheta-%5Cpi%29%3D%5Cfrac%7B%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D)
d)![\tan(\theta + \frac{\pi}{4}) = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%20%2B%20%5Cfrac%7B%5Cpi%7D%7B4%7D%29%20%3D%20%5Cfrac%7B%5Cfrac%7B-4%7D%7B%5Csqrt%5B%5D%7B33%7D%7D%2B1%7D%7B1%2B%5Cfrac%7B4%7D%7B%5Csqrt%5B%5D%7B33%7D%7D%7D)
Step-by-step explanation:
We will use the following trigonometric identities
.
Recall that given a right triangle, the sin(theta) is defined by opposite side/hypotenuse. Since we know that the angle is in quadrant 2, we know that x should be a negative number. We will use pythagoras theorem to find out the value of x. We have that
which implies that ![x=-\sqrt[]{49-16} = -\sqrt[]{33}](https://tex.z-dn.net/?f=x%3D-%5Csqrt%5B%5D%7B49-16%7D%20%3D%20-%5Csqrt%5B%5D%7B33%7D)
. Recall that cos(theta) is defined by adjacent side/hypotenuse. So, we know that the hypotenuse is 7, then
![\cos(\theta) = \frac{-\sqrt[]{33}}{7}](https://tex.z-dn.net/?f=%5Ccos%28%5Ctheta%29%20%3D%20%5Cfrac%7B-%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D)
b)Recall that ![\sin(\frac{\pi}{6}) =\frac{1}{2} , \cos(\frac{\pi}{6}) = \frac{\sqrt[]{3}}{2}](https://tex.z-dn.net/?f=%5Csin%28%5Cfrac%7B%5Cpi%7D%7B6%7D%29%20%3D%5Cfrac%7B1%7D%7B2%7D%20%2C%20%5Ccos%28%5Cfrac%7B%5Cpi%7D%7B6%7D%29%20%3D%20%5Cfrac%7B%5Csqrt%5B%5D%7B3%7D%7D%7B2%7D)
, then using the identity from above, we have that
![\sin(\theta + \frac{\pi}{6}) = \sin(\theta)\cos(\frac{\pi}{6})+\cos(\alpha)\sin(\frac{\pi}{6}) = \frac{4}{7}\frac{1}{2}-\frac{\sqrt[]{33}}{7}\frac{\sqrt[]{3}}{2} = \frac{-3\sqrt[]{11}+4}{14}](https://tex.z-dn.net/?f=%5Csin%28%5Ctheta%20%2B%20%5Cfrac%7B%5Cpi%7D%7B6%7D%29%20%3D%20%5Csin%28%5Ctheta%29%5Ccos%28%5Cfrac%7B%5Cpi%7D%7B6%7D%29%2B%5Ccos%28%5Calpha%29%5Csin%28%5Cfrac%7B%5Cpi%7D%7B6%7D%29%20%3D%20%5Cfrac%7B4%7D%7B7%7D%5Cfrac%7B1%7D%7B2%7D-%5Cfrac%7B%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D%5Cfrac%7B%5Csqrt%5B%5D%7B3%7D%7D%7B2%7D%20%3D%20%5Cfrac%7B-3%5Csqrt%5B%5D%7B11%7D%2B4%7D%7B14%7D)
c) Recall that 
. Then,
![\cos(\theta-\pi)=\cos(\theta)\cos(\pi)+\sin(\theta)\sin(\pi) = \frac{-\sqrt[]{33}}{7}\cdot(-1) + 0 = \frac{\sqrt[]{33}}{7}](https://tex.z-dn.net/?f=%5Ccos%28%5Ctheta-%5Cpi%29%3D%5Ccos%28%5Ctheta%29%5Ccos%28%5Cpi%29%2B%5Csin%28%5Ctheta%29%5Csin%28%5Cpi%29%20%3D%20%5Cfrac%7B-%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D%5Ccdot%28-1%29%20%2B%200%20%3D%20%5Cfrac%7B%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D)
d) Recall that 
and ![\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}=\frac{-4}{\sqrt[]{33}}](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%29%20%3D%20%5Cfrac%7B%5Csin%28%5Ctheta%29%7D%7B%5Ccos%28%5Ctheta%29%7D%3D%5Cfrac%7B-4%7D%7B%5Csqrt%5B%5D%7B33%7D%7D)
. Then
![\tan(\theta+\frac{\pi}{4}) = \frac{\tan(\theta)+\tan(\frac{\pi}{4})}{1-\tan(\theta)\tan(\frac{\pi}{4})} = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%2B%5Cfrac%7B%5Cpi%7D%7B4%7D%29%20%3D%20%5Cfrac%7B%5Ctan%28%5Ctheta%29%2B%5Ctan%28%5Cfrac%7B%5Cpi%7D%7B4%7D%29%7D%7B1-%5Ctan%28%5Ctheta%29%5Ctan%28%5Cfrac%7B%5Cpi%7D%7B4%7D%29%7D%20%3D%20%5Cfrac%7B%5Cfrac%7B-4%7D%7B%5Csqrt%5B%5D%7B33%7D%7D%2B1%7D%7B1%2B%5Cfrac%7B4%7D%7B%5Csqrt%5B%5D%7B33%7D%7D%7D)
Hi!
The answer is 6.2664 feet!
Hope this helps:))
Answer:
I need to see the choices for it
Step-by-step explanation:
Answer:
The area of the parallelogram is 
Step-by-step explanation:
Remember that a mixed fractions are composed for a whole number, a numerator and a denominator. For example in the mixed fraction 
: 2 is the whole number, 1 is the numerator and 3 is the denominator.
If you want to convert to a improper fraction you need to multiply the whole number by the denominator and add to the numerator. The result will be the numerator of the improper fraction. The denominator remains the same.
1. Convert the mixed fractions in the improper fraction type, as follows:
2. Replace the data in the equation of a parallelogram
3. Convert the improper fraction in a mixed fraction
The area of the parallelogram is
