Answer:
![\sin(105) = \frac{\sqrt 2 + \sqrt 6}{4}](https://tex.z-dn.net/?f=%5Csin%28105%29%20%3D%20%5Cfrac%7B%5Csqrt%202%20%2B%20%5Csqrt%206%7D%7B4%7D)
Step-by-step explanation:
Given
![\sin(105^o)](https://tex.z-dn.net/?f=%5Csin%28105%5Eo%29)
Required
Solve
Using sine rule, we have:
![\sin(A + B) = \sin(A)\cos(B) + \sin(B)\cos(A)](https://tex.z-dn.net/?f=%5Csin%28A%20%2B%20B%29%20%3D%20%5Csin%28A%29%5Ccos%28B%29%20%2B%20%5Csin%28B%29%5Ccos%28A%29)
This gives:
![\sin(105^o) = \sin(60 + 45)](https://tex.z-dn.net/?f=%5Csin%28105%5Eo%29%20%3D%20%5Csin%2860%20%2B%2045%29)
So, we have:
![\sin(60 + 45) = \sin(60)\cos(45) + \sin(45)\cos(60)](https://tex.z-dn.net/?f=%5Csin%2860%20%2B%2045%29%20%3D%20%5Csin%2860%29%5Ccos%2845%29%20%2B%20%5Csin%2845%29%5Ccos%2860%29)
In radical forms, we have:
![\sin(60 + 45) = \frac{\sqrt 3}{2} * \frac{\sqrt 2}{2} + \frac{\sqrt 2}{2} * \frac{1}{2}](https://tex.z-dn.net/?f=%5Csin%2860%20%2B%2045%29%20%3D%20%5Cfrac%7B%5Csqrt%203%7D%7B2%7D%20%2A%20%5Cfrac%7B%5Csqrt%202%7D%7B2%7D%20%2B%20%5Cfrac%7B%5Csqrt%202%7D%7B2%7D%20%2A%20%5Cfrac%7B1%7D%7B2%7D)
![\sin(60 + 45) = \frac{\sqrt 6}{4} + \frac{\sqrt 2}{4}](https://tex.z-dn.net/?f=%5Csin%2860%20%2B%2045%29%20%3D%20%5Cfrac%7B%5Csqrt%206%7D%7B4%7D%20%2B%20%5Cfrac%7B%5Csqrt%202%7D%7B4%7D)
Take LCM
![\sin(60 + 45) = \frac{\sqrt 6 + \sqrt 2}{4}](https://tex.z-dn.net/?f=%5Csin%2860%20%2B%2045%29%20%3D%20%5Cfrac%7B%5Csqrt%206%20%2B%20%5Csqrt%202%7D%7B4%7D)
Rewrite as:
![\sin(60 + 45) = \frac{\sqrt 2 + \sqrt 6}{4}](https://tex.z-dn.net/?f=%5Csin%2860%20%2B%2045%29%20%3D%20%5Cfrac%7B%5Csqrt%202%20%2B%20%5Csqrt%206%7D%7B4%7D)
Hence:
![\sin(105) = \frac{\sqrt 2 + \sqrt 6}{4}](https://tex.z-dn.net/?f=%5Csin%28105%29%20%3D%20%5Cfrac%7B%5Csqrt%202%20%2B%20%5Csqrt%206%7D%7B4%7D)
The answer to this is 384.57909
Maximum area of the rectangle is ![9cm^{2}](https://tex.z-dn.net/?f=9cm%5E%7B2%7D)
<u>Explanation:</u>
<u></u>
Considering the dimensions to be in cm
![f(x) = -(x-3)^{2} +9\\f(x) = -(x^{2} +9 - 6x)+9\\f(x) = -x^{2} +6x\\f'(x) = -2x+6\\-2x+6 = 0\\2x=6\\x=3cm\\\\](https://tex.z-dn.net/?f=f%28x%29%20%3D%20-%28x-3%29%5E%7B2%7D%20%2B9%5C%5Cf%28x%29%20%3D%20-%28x%5E%7B2%7D%20%2B9%20-%206x%29%2B9%5C%5Cf%28x%29%20%3D%20-x%5E%7B2%7D%20%2B6x%5C%5Cf%27%28x%29%20%3D%20-2x%2B6%5C%5C-2x%2B6%20%3D%200%5C%5C2x%3D6%5C%5Cx%3D3cm%5C%5C%5C%5C)
Putting the value of x = 3
![Perimeter = 2(x+b)\\12 = 2(3+b)\\6 = 3+b\\b= 3cm](https://tex.z-dn.net/?f=Perimeter%20%3D%202%28x%2Bb%29%5C%5C12%20%3D%202%283%2Bb%29%5C%5C6%20%3D%203%2Bb%5C%5Cb%3D%203cm)
Therefore, maximum area of the rectangle is ![9cm^{2}](https://tex.z-dn.net/?f=9cm%5E%7B2%7D)
Answer:
y = -3x - 1
Step-by-step explanation:
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Answer:
WHERE ARE THE PROBLEMS/VISUALS
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Step-by-step explanation: