Question

note: images are not drawn to scale.
what is \\(\sin d\\)?
\\(\sin d = \boxed{\frac{3}{5}}\\)

Explanation:

Step1: Identify Triangle Similarity

Triangles \(ABC\) and \(DFE\) are right - angled triangles. Since they are both right - angled, if they are similar, their corresponding angles are equal. For right - angled triangles, if the ratios of corresponding sides are equal, the triangles are similar. In \(\triangle ABC\), \(AC = 9\), \(BC=12\), \(AB = 15\). In \(\triangle DFE\), \(\angle F=90^{\circ}\) (same as \(\angle C = 90^{\circ}\)). So, \(\triangle ABC\sim\triangle DEF\) (by AA similarity criterion, as both are right - angled and the other angles will be equal because of the right angle). So, \(\angle D=\angle A\) or \(\angle D\) has the same sine as the corresponding angle in \(\triangle ABC\).

Step2: Recall Sine Definition

The sine of an angle in a right - triangle is defined as \(\sin\theta=\frac{\text{opposite side}}{\text{hypotenuse}}\). In \(\triangle ABC\), for angle \(A\), the opposite side to \(\angle A\) is \(BC = 12\), and the hypotenuse is \(AB=15\)? Wait, no. Wait, in \(\triangle ABC\), \(\angle C = 90^{\circ}\), so for \(\angle A\), opposite side is \(BC = 12\), adjacent is \(AC = 9\), hypotenuse \(AB = 15\). But for \(\angle B\), opposite side is \(AC=9\), hypotenuse \(AB = 15\). Wait, since \(\triangle ABC\sim\triangle DEF\), \(\angle D\) corresponds to \(\angle A\) or \(\angle B\)? Wait, let's re - examine. In \(\triangle ABC\), sides: \(AC = 9\), \(BC = 12\), \(AB=15\). In \(\triangle DFE\), right - angled at \(F\). So, \(\sin D=\frac{\text{opposite side to }D}{\text{hypotenuse of }\triangle DFE}\). Since the triangles are similar, the ratio of sides is the same as in \(\triangle ABC\). Let's find \(\sin A\) or \(\sin B\). Wait, in \(\triangle ABC\), \(\sin B=\frac{AC}{AB}=\frac{9}{15}=\frac{3}{5}\). Since \(\triangle ABC\sim\triangle DEF\), \(\angle D=\angle B\) (because of the correspondence of similar triangles). So, \(\sin D=\sin B=\frac{3}{5}\).

Answer:

\(\frac{3}{5}\)