In triangle ABC , AB = x cm , BC = (4-x) cm , angle BAC = y degrees and angle BCA = 30 degrees.
Given that sin y = , show that
what i got is:
where did i go wrong ?
Law of sines ...
$\dfrac{x}{\sin(30)} = \dfrac{4-x}{\sin{y}}$
$2x = \sqrt{2}(4-x)$
$2x = 4\sqrt{2} - x\sqrt{2}$
$2x + x\sqrt{2} = 4\sqrt{2}$
$x(2 + \sqrt{2}) = 4\sqrt{2}$
$x = \dfrac{4\sqrt{2}}{2+\sqrt{2}} \cdot \dfrac{2-\sqrt{2}}{2-\sqrt{2}} = \dfrac{8\sqrt{2}-8}{2} = 4\sqrt{2} - 4 = 4(\sqrt{2}-1)$