Question

in the diagram of right triangle edc below, altitude \\(\overline{db}\\) is drawn. which of the following ratios is equivalent to \\(\sin c\\)?
answer
\\(\frac{db}{dc}\\) \\(\frac{dc}{ec}\\)
\\(\frac{bc}{db}\\) \\(\frac{eb}{bd}\\)

Explanation:

Step1: Define sin C in △EDC

In right triangle $EDC$ (right-angled at $D$), $\sin C = \frac{\text{opposite side to } C}{\text{hypotenuse}} = \frac{ED}{EC}$

Step2: Identify similar triangles

Since $DB$ is an altitude, $\triangle CDB \sim \triangle CED$ and $\triangle EDB \sim \triangle ECD$. Also, $\angle C = \angle EDB$ (corresponding angles of similar triangles).

Step3: Find sin(∠EDB)

In right triangle $EDB$, $\sin(\angle EDB) = \frac{EB}{ED}$. But we can also use the similarity of $\triangle CDB$ and $\triangle CED$: $\sin C = \frac{DB}{BC}$ is incorrect. Instead, in $\triangle CDB$, $\sin C = \frac{DB}{BC}$ is wrong; correct: in $\triangle EDC$, $\sin C = \frac{ED}{EC}$, and from $\triangle EDB \sim \triangle ECD$, $\frac{ED}{EC} = \frac{EB}{ED}$, but also, in $\triangle CDB$, $\sin C = \frac{DB}{BC}$ is wrong. Wait, correct approach: In right $\triangle EDC$, $\sin C = \frac{ED}{EC}$. In right $\triangle EDB$, $\sin(\angle EDB) = \frac{EB}{ED}$, but $\angle C = \angle EDB$, so $\sin C = \sin(\angle EDB) = \frac{EB}{BD}$. Wait no, let's recheck:
In $\triangle EDC$, right at $D$, $\angle C + \angle E = 90^\circ$. In $\triangle EDB$, right at $B$, $\angle E + \angle EDB = 90^\circ$, so $\angle C = \angle EDB$. Then $\sin C = \sin(\angle EDB) = \frac{EB}{ED}$? No, in $\triangle EDB$, right angle at $B$, so $\sin(\angle EDB) = \frac{EB}{ED}$. But also, in $\triangle CDB$, right angle at $B$, $\sin C = \frac{DB}{DC}$? No, wait:
Wait, in right $\triangle EDC$, $\sin C = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{ED}{EC}$.
From similar triangles $\triangle EDB \sim \triangle ECD$: $\frac{ED}{EC} = \frac{EB}{ED} = \frac{DB}{DC}$. Also, $\triangle CDB \sim \triangle CED$: $\frac{CD}{CE} = \frac{CB}{CD} = \frac{DB}{ED}$.
Wait, let's do it properly:
In $\triangle CDB$, right-angled at $B$, $\sin C = \frac{DB}{DC}$? No, $\sin C = \frac{\text{opposite side to } C}{\text{hypotenuse}} = \frac{DB}{DC}$? Wait, hypotenuse of $\triangle CDB$ is $DC$, opposite side to $C$ is $DB$. Yes! Wait no, $\triangle CDB$: right angle at $B$, so angle at $C$, opposite side is $DB$, hypotenuse is $DC$. So $\sin C = \frac{DB}{DC}$? No, wait no: $\triangle EDC$ is right-angled at $D$, so angle $C$ is at vertex $C$, sides: $ED$ is opposite, $EC$ is hypotenuse, $DC$ is adjacent. So $\sin C = \frac{ED}{EC}$.
Now, $\triangle EDB \sim \triangle ECD$ (AA similarity: $\angle E$ is common, right angles $\angle EBD = \angle EDC$). So $\frac{ED}{EC} = \frac{EB}{ED} = \frac{DB}{DC}$. So $\sin C = \frac{ED}{EC} = \frac{DB}{DC}$. Wait no, $\frac{DB}{DC}$ is $\cos C$ in $\triangle CDB$. Oh right! I messed up. In $\triangle CDB$, right at $B$, $\cos C = \frac{CB}{DC}$, $\sin C = \frac{DB}{DC}$? No, no: in $\triangle CDB$, angle at $C$, adjacent side is $CB$, opposite side is $DB$, hypotenuse is $DC$. So $\sin C = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{DB}{DC}$? But in $\triangle EDC$, $\sin C = \frac{ED}{EC}$. Are these equal?
Yes, because $\triangle EDC \sim \triangle DBC$ (AA: $\angle C$ is common, right angles $\angle EDC = \angle DBC$). So $\frac{ED}{DB} = \frac{EC}{DC} = \frac{DC}{BC}$. So $\frac{ED}{EC} = \frac{DB}{DC}$. Therefore, $\sin C = \frac{ED}{EC} = \frac{DB}{DC}$.
Wait, but also, from $\triangle EDB \sim \triangle ECD$, $\frac{EB}{ED} = \frac{ED}{EC}$, so $\frac{ED}{EC} = \frac{EB}{ED}$, but that's not an option. The options are $\frac{DB}{DC}$, $\frac{DC}{EC}$, $\frac{BC}{DB}$, $\frac{EB}{BD}$.
Wait, let's check $\frac{EB}{BD}$: in $\triangle EDB$, $\tan(\angle EDB) = \frac{EB}{BD}$, and $\angle EDB = \angle C$, so $\tan C = \frac{EB}{BD}$, not $\sin C$.
$\frac{BC}{DB}$: in $\triangle CDB$, $\tan C = \frac{DB}{BC}$, so $\frac{BC}{DB} = \cot C$.
$\frac{DC}{EC}$: that's $\cos C$, since in $\triangle EDC$, $\cos C = \frac{DC}{EC}$.
$\frac{DB}{DC}$: that's $\sin C$, as we saw from similarity.
Wait no, wait in $\triangle EDC$, $\sin C = \frac{ED}{EC}$, and from $\triangle EDC \sim \triangle DBC$, $\frac{ED}{DB} = \frac{EC}{DC}$, so $\frac{ED}{EC} = \frac{DB}{DC}$. Yes, so $\sin C = \frac{DB}{DC}$.

Wait, I made a mistake earlier: in $\triangle CDB$, $\sin C = \frac{DB}{DC}$ is correct, because opposite side to $C$ is $DB$, hypotenuse is $DC$. Yes, that's right. Because $\triangle CDB$ is right-angled at $B$, so hypotenuse is $DC$, angle at $C$, opposite side is $DB$. So $\sin C = \frac{DB}{DC}$.

Answer:

$\boldsymbol{\frac{DB}{DC}}$ (Option A)