Question

what is the value of x in the figure shown?

Explanation:

Step1: Find the length of \(AC\)

In right - triangle \(ABC\) with \(\angle B = 30^{\circ}\), we use the property that in a \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle, if the side opposite the \(30^{\circ}\) angle is \(x\), the hypotenuse is \(2x\) and the side opposite the \(60^{\circ}\) angle is \(x\sqrt{3}\). Here, \(AB\) is the hypotenuse, \(AC\) is the side opposite \(\angle B\). Given \(BC = 2\sqrt{3}\), and \(AC=\frac{1}{2}AB\) (by \(30 - 60-90\) triangle property: \(\sin B=\frac{AC}{AB}\), \(\sin30^{\circ}=\frac{AC}{AB}=\frac{1}{2}\)), also using \(\cos B=\frac{BC}{AB}\), \(AB = 4\) (since \(\cos30^{\circ}=\frac{BC}{AB}\), \(AB=\frac{BC}{\cos30^{\circ}}=\frac{2\sqrt{3}}{\frac{\sqrt{3}}{2}} = 4\)), and \(AC = 2\) (because \(AC=\frac{1}{2}AB\)).

Step2: Find the value of \(x\)

In right - triangle \(ACD\) with \(\angle A=60^{\circ}\), using the property of a \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle (\(\cos A=\frac{AD}{AC}\)). Since \(\angle A = 60^{\circ}\) and \(AC = 2\), and \(\cos60^{\circ}=\frac{x}{AC}\). Substitute \(AC = 2\) into the formula \(\cos60^{\circ}=\frac{x}{2}\). Since \(\cos60^{\circ}=\frac{1}{2}\), we have \(x = 1\).

Answer:

\(1\)