We know that sum of the interior angles of any triangle equal 180° :
So :
A + B + C = 180°
45° + 75° + C = 180°
120° + C = 180°
Both sides minus ( 120°) :
C = 180° - 120°
C = 60°
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According to the theorem of sinuses :
[tex] \frac{AB}{ \sin(C) } = \frac{BC}{ \sin(A) } \\ [/tex]
[tex] \frac{AB}{ \sin(60) } = \frac{12}{ \sin(45) } \\ [/tex]
[tex] \frac{AB}{ \frac{ \sqrt{3} }{2} } = \frac{12}{ \frac{ \sqrt{2} }{2} } \\ [/tex]
[tex]Multiply \: both \: sides \: by \: \frac{ \sqrt{3} }{2} [/tex]
[tex] AB = \frac{24}{ \sqrt{2} } \times \frac{ \sqrt{3} }{2} \\ [/tex]
[tex]AB = \frac{12 \sqrt{3} }{ \sqrt{2} } \\ [/tex]
[tex]AB = \frac{2 \times 6 \sqrt{3} }{ \sqrt{2} } \\ [/tex]
[tex]AB = \frac{ ({ \sqrt{2} })^{2} \times 6 \sqrt{3} }{ \sqrt{2} } \\ [/tex]
[tex]AB = \frac{ \sqrt{2} \times \sqrt{2} \times 6 \sqrt{3} }{ \sqrt{2} } \\ [/tex]
[tex]AB = \frac{ \sqrt{2} \times 6 \sqrt{3} }{1} \\ [/tex]
[tex]AB = \sqrt{2} \times 6 \sqrt{3} [/tex]
[tex]AB = 6 \sqrt{2 \times 3} [/tex]
[tex]AB = 6 \sqrt{6} [/tex]
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I think this is the correct answer.
And we're done.
Thanks for watching buddy good luck.
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