Question

in the diagram, \\( \triangle xyz \\) is an equilateral triangle. \\( \overline{wy} \\) bisects \\( \angle xyz \\) and \\( \overline{xz} \\). find to the nearest tenth the length of a side of \\( \triangle xyz \\) if \\( wy = 23 \\).

Explanation:

Step1: Analyze the properties of the equilateral triangle

Since \(\triangle XYZ\) is equilateral, \(\angle XYZ = 60^{\circ}\). \(WY\) bisects \(\angle XYZ\), so \(\angle WYZ=\frac{1}{2}\angle XYZ = 30^{\circ}\). Also, \(WY\) bisects \(XZ\), so \(WY\perp XZ\) (by the property of an equilateral triangle: the angle - bisector, median, and altitude from a vertex to the opposite side are the same).

Step2: Use trigonometric ratios

In right - triangle \(WYZ\), let the side length of \(\triangle XYZ\) be \(s\), then \(YZ = s\) and \(WZ=\frac{s}{2}\). Using the cosine ratio \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), and \(\theta = 30^{\circ}\), \(\cos30^{\circ}=\frac{WZ}{YZ}\). But we can also use the relationship in a \(30 - 60-90\) triangle. In a right - triangle with angles \(30^{\circ}-60^{\circ}-90^{\circ}\), if the side opposite the \(30^{\circ}\) angle is \(a\), the side opposite the \(60^{\circ}\) angle is \(a\sqrt{3}\), and the hypotenuse is \(2a\). Here, if we consider \(\triangle WYZ\) (right - angled at \(W\)), and let \(WZ = x\), \(YZ = 2x\), and \(WY=\sqrt{3}x\). Given \(WY = 23\), then \(\sqrt{3}x=23\), \(x=\frac{23}{\sqrt{3}}\). The side length of the equilateral triangle \(s = 2x\).

$$s=\frac{46}{\sqrt{3}}\approx\frac{46}{1.732}\approx26.6$$

Answer:

\(26.6\)