Question

triangles abc and def are right triangles, as shown. triangle abc is similar to triangle def. which ratios are equal to sin c? select all that apply. a $\frac{ab}{ac}$ b $\frac{ab}{bc}$ c $\frac{bc}{ac}$ d $\frac{de}{df}$ e $\frac{de}{ef}$ f $\frac{ef}{df}$

Explanation:

Step1: Recall sine in right triangle

In right triangle \(ABC\) (right - angled at \(B\)), \(\sin C=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{AB}{AC}\).

Step2: Use similarity of triangles

Since \(\triangle ABC\sim\triangle DEF\) (right - angled at \(B\) and \(E\) respectively), corresponding angles are equal. So \(\angle C=\angle F\) (corresponding angles of similar triangles).
In right triangle \(DEF\) (right - angled at \(E\)), \(\sin F = \frac{DE}{DF}\) (because for \(\angle F\), the opposite side is \(DE\) and the hypotenuse is \(DF\)). Also, in \(\triangle ABC\), \(\sin C=\frac{AB}{AC}\), and due to similarity, the ratios of corresponding sides are equal. So \(\frac{AB}{AC}=\frac{DE}{DF}\) and \(\sin C=\sin F=\frac{DE}{DF}\). Also, in \(\triangle ABC\), \(\sin C = \frac{AB}{AC}\), and let's check other options:

• Option B: \(\frac{AB}{BC}\) is not \(\sin C\) (since \(\sin C=\frac{AB}{AC}\), \(BC\) is adjacent to \(\angle C\) in right - triangle \(ABC\)).
• Option C: \(\frac{BC}{AC}\) is \(\cos C\) (adjacent over hypotenuse), not \(\sin C\).
• Option E: \(\frac{DE}{EF}\) is not equal to \(\sin C\) (in \(\triangle DEF\), \(EF\) is adjacent to \(\angle F\) (which is equal to \(\angle C\)), so \(\frac{DE}{EF}=\tan F=\tan C\), not \(\sin C\)).
• Option F: \(\frac{EF}{DF}\) is \(\cos F=\cos C\), not \(\sin C\).

Answer:

A. \(\boldsymbol{\frac{AB}{AC}}\), D. \(\boldsymbol{\frac{DE}{DF}}\)