Respuesta :

[tex]~~~~~~~\cos 45^{\circ} = \dfrac{\text{Base}}{\text{Hypotenuse}}\\\\\\\implies \dfrac 1{\sqrt 2} = \dfrac{\sqrt 3}{x}\\\\\\\implies x = \sqrt 3 \cdot \sqrt 2\\\\\\\implies x = \sqrt {3 \times 2}\\\\\\\implies x = \sqrt 6[/tex]

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