1. sine rule

In triangle ABC , AB = x cm , BC = (4-x) cm , angle BAC = y degrees and angle BCA = 30 degrees.
Given that sin y = $\frac{1}{\sqrt{2}}$, show that $x=4(\sqrt{2}-1)$

what i got is:

$\frac{(4-x)}{\sin y}=\frac{x}{\sin 30}\\x=2\sqrt{2}-\frac{\sqrt{2}x}{2}\\2x=4\sqrt{2}-\sqrt{2}x\\x(2-\sqrt{2})=4\sqrt{2}\\x=2\sqrt{2}-4\\x=2(\sqrt{2}-2)$

where did i go wrong ?

2. Re: sine rule

Law of sines ...

$\dfrac{x}{\sin(30)} = \dfrac{4-x}{\sin{y}}$

$2x = \sqrt{2}(4-x)$

$2x = 4\sqrt{2} - x\sqrt{2}$

$2x + x\sqrt{2} = 4\sqrt{2}$

$x(2 + \sqrt{2}) = 4\sqrt{2}$

$x = \dfrac{4\sqrt{2}}{2+\sqrt{2}} \cdot \dfrac{2-\sqrt{2}}{2-\sqrt{2}} = \dfrac{8\sqrt{2}-8}{2} = 4\sqrt{2} - 4 = 4(\sqrt{2}-1)$

3. Re: sine rule

Originally Posted by AbYz
In triangle ABC , AB = x cm , BC = (4-x) cm , angle BAC = y degrees and angle BCA = 30 degrees.
Given that sin y = $\frac{1}{\sqrt{2}}$, show that $x=4(\sqrt{2}-1)$

what i got is:

$\frac{(4-x)}{\siny}=\frac{x}{\sin30}$
$x=2\sqrt{2}\-\frac{sqrt{2}x}{2}$ $x=2\sqrt{2}-4$ $x=2(\sqrt{2}-2)$

where did i go wrong ?
$sin(30)= 1/2$, not $\sqrt{2}/2$