The double angle formula for the sine is

The right hand side is exactly your expression, where 
So, rewriting the expression from right to left, we have

 
        
             
        
        
        
We can find out what LM is based on the fact that the bases of the 2 triangles are the same, this would also mean that the hypotenuse sides of the triangles should also be the same as well. Thus the sides of LM should also equal the sides of MN. 
Another way, is to assume that the triangles are right angle ones, and use Pythagorean theorem to solve for the height and use that to solve for the hypotenuse. 
        
             
        
        
        
Answer:
 49+0.89c
Step-by-step explanation:
This should be the answer good luck :)
 
        
             
        
        
        
Answer:
![W=\{\left[\begin{array}{ccc}a+2b\\b\\-3a\end{array}\right]: a,b\in\mathbb{R} \}](https://tex.z-dn.net/?f=W%3D%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%2B2b%5C%5Cb%5C%5C-3a%5Cend%7Barray%7D%5Cright%5D%3A%20a%2Cb%5Cin%5Cmathbb%7BR%7D%20%5C%7D)
Observe that if the vector ![x=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]](https://tex.z-dn.net/?f=x%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D) is in W then it satisfies:
 is in W then it satisfies:
![\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{c}a+2b\\b\\-3a\end{array}\right]=a\left[\begin{array}{c}1\\0\\-3\end{array}\right]+b\left[\begin{array}{c}2\\1\\0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Da%2B2b%5C%5Cb%5C%5C-3a%5Cend%7Barray%7D%5Cright%5D%3Da%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%5C%5C0%5C%5C-3%5Cend%7Barray%7D%5Cright%5D%2Bb%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C1%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
This means that each vector in W can be expressed as a linear combination of the vectors ![\left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}2\\1\\0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%5C%5C0%5C%5C-3%5Cend%7Barray%7D%5Cright%5D%2C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C1%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
Also we can see that those vectors are linear independent. Then the set 
![\{\left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}2\\1\\0\end{array}\right]\}](https://tex.z-dn.net/?f=%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%5C%5C0%5C%5C-3%5Cend%7Barray%7D%5Cright%5D%2C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C1%5C%5C0%5Cend%7Barray%7D%5Cright%5D%5C%7D) is a basis for W and the dimension of W is 2.
 is a basis for W and the dimension of W is 2.
 
        
             
        
        
        
Answer:
s=x/3+10/3
Step-by-step explanation: