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AOBC is a parallelogram ,
Thus :
[tex]OA = BC = a[/tex]
And ;
[tex]AC = OB = b [/tex]
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M is the midpoint of AC ,
So :
[tex]AM = MC = \frac{b}{2} \\ [/tex]
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According to case of Cosine in ∆ OAM :
[tex] {OM}^{2} = {OA}^{2} + {AM}^{2} - 2( OA) \times ( AM) \times \cos(A) \\ [/tex]
So :
[tex] {OM}^{2} = {a}^{2} + \frac{ {b}^{2} }{4} - 2(a)( \frac{b}{2} ) \cos(A) \\ [/tex]
[tex] {OM}^{2} = \frac{4 {a}^{2} + b}{4} - ab \cos(A) \\ [/tex]
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According to the case of Sine in ∆ OAM :
[tex] \frac{OM}{ sin(A) } = \frac{
AM}{ \sin(O) } = \frac{OA}{ \sin(M) } \\ [/tex]
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