Question

in the diagram of right triangle vut below, altitude \\(\overline{us}\\) is drawn. which of the following ratios is equivalent to \\(\tan t\\)?

answer
\\(\frac{ut}{vu}\\) \\(\frac{st}{ut}\\) \\(\frac{vs}{vu}\\) \\(\frac{vs}{su}\\)

Explanation:

Step1: Recall tan definition

In right triangle \( VUT \) (right-angled at \( U \)), \( \tan T=\frac{\text{opposite}}{\text{adjacent}}=\frac{VU}{UT} \). Also, \( \overline{US} \) is an altitude, so triangles \( VUS \), \( UST \), and \( VUT \) are similar.

Step2: Analyze similar triangles

For \( \triangle UST \) (right-angled at \( S \)) and \( \triangle VUT \), \( \angle T \) is common. So \( \tan T \) in \( \triangle UST \) is \( \frac{SU}{ST} \), but also, in \( \triangle VUS \) (right-angled at \( S \)), \( \tan T=\tan \angle VUS \) (since \( \angle VUS = \angle T \) from similarity). Wait, another approach: check the ratios. Let's see the options. The option \( \frac{VS}{SU} \): Wait, no, let's re-express. Wait, in \( \triangle VUS \) and \( \triangle UST \), since \( \triangle VUT \sim \triangle UST \sim \triangle VUS \). So \( \tan T=\frac{VU}{UT}=\frac{SU}{ST}=\frac{VS}{SU} \)? Wait, let's check the last option \( \frac{VS}{SU} \). Wait, let's verify:

In \( \triangle VUS \), right-angled at \( S \), \( \tan \angle VUS=\frac{VS}{SU} \). But \( \angle VUS = \angle T \) (because \( \angle VUS + \angle UVS = 90^\circ \), and \( \angle T + \angle UVS = 90^\circ \), so they are equal). Therefore, \( \tan T = \tan \angle VUS=\frac{VS}{SU} \). So the correct ratio is \( \frac{VS}{SU} \), which is the last option (the fourth option, \( \frac{VS}{SU} \)). Wait, let's check the options again:

Options:

1. \( \frac{UT}{VU} \) – reciprocal of \( \tan T \), no.
2. \( \frac{VS}{VU} \) – no, that's not.
3. \( \frac{ST}{UT} \) – no, \( \tan T=\frac{SU}{ST} \), so \( \frac{ST}{UT} \) is not.
4. \( \frac{VS}{SU} \) – yes, because \( \angle VUS = \angle T \), so \( \tan T = \tan \angle VUS=\frac{VS}{SU} \) (opposite over adjacent in \( \triangle VUS \)).

Wait, maybe I made a mistake earlier. Let's re-express:

In \( \triangle VUT \), right-angled at \( U \), \( \tan T = \frac{VU}{UT} \).

In \( \triangle UST \), right-angled at \( S \), \( \tan T = \frac{SU}{ST} \).

In \( \triangle VUS \), right-angled at \( S \), \( \tan \angle VUS = \frac{VS}{SU} \). But \( \angle VUS = \angle T \) (since \( \angle VUS + \angle UVS = 90^\circ \) and \( \angle T + \angle UVS = 90^\circ \)), so \( \tan T = \tan \angle VUS = \frac{VS}{SU} \). Therefore, the ratio \( \frac{VS}{SU} \) is equivalent to \( \tan T \).

Answer:

\(\boldsymbol{\frac{VS}{SU}}\) (the fourth option, i.e., the option with ratio \(\frac{VS}{SU}\))