Show that \(\sin 105^\circ=\dfrac{\sqrt{6}+\sqrt{2}}{4}\).[3]
The diagram below shows triangle \(ABC\), such that \(AB=4\) cm, angle \(ABC=105^\circ\) and the area of triangle \(ABC\) is \(10+2\sqrt{3}\) cm\(^2\).[3]
Find \(BC\) in the form \(\sqrt{2}(a\sqrt{3}+b)\), where \(a\) and \(b\) are integers.
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