Question

in the diagram of right triangle fed below, altitude \\(\overline{ec}\\) is drawn. which of the following ratios is equivalent to \\(\sin d\\)?

diagram of right triangle fed with right angle at e, and altitude ec drawn from e to hypotenuse fd, meeting at point c

answer
\\(\circ\\) \\(\frac{fc}{ce}\\)
\\(\circ\\) \\(\frac{cd}{ec}\\)
\\(\circ\\) \\(\frac{ed}{fd}\\)
\\(\circ\\) \\(\frac{fe}{fd}\\)

Explanation:

Step1: Recall Sine Definition

In right triangle \( FED \) (right - angled at \( E \)), \( \sin D=\frac{\text{opposite side to }D}{\text{hypotenuse}} \). The side opposite to \( \angle D \) is \( FE \), and the hypotenuse is \( FD \). So, \( \sin D = \frac{FE}{FD} \).

Step2: Analyze Similar Triangles (Optional)

Since \( EC \) is an altitude to the hypotenuse of right triangle \( FED \), triangles \( FED \), \( FEC \), and \( ECD \) are similar. In triangle \( FED \), \( \sin D=\frac{FE}{FD} \). Let's check other options:

• For \( \frac{FC}{CE} \): In triangle \( FEC \), \( \tan\angle F=\frac{FC}{CE} \), not related to \( \sin D \).
• For \( \frac{CD}{EC} \): In triangle \( ECD \), \( \tan D=\frac{CD}{EC} \), not \( \sin D \).
• For \( \frac{ED}{FD} \): \( \cos D=\frac{ED}{FD} \), not \( \sin D \).

Answer:

\( \boldsymbol{\frac{FE}{FD}} \) (the option with \( \frac{FE}{FD} \))