Question

in the diagram of right triangle srq below, altitude \\(\overline{rp}\\) is drawn. which of the following ratios is equivalent to \\(\cos q\\)?

answer
\\(\bigcirc\\) \\(\frac{rp}{rq}\\) \\(\bigcirc\\) \\(\frac{sr}{rq}\\)
\\(\bigcirc\\) \\(\frac{pr}{sp}\\) \\(\bigcirc\\) \\(\frac{pq}{rq}\\)

Explanation:

Step1: Recall cosine definition

In a right triangle, $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$. For $\angle Q$ in $\triangle SRQ$ (right-angled at $R$), $\cos Q = \frac{RQ}{SQ}$? Wait, no, wait. Wait, $\triangle SRQ$ is right-angled at $R$, so $\angle R = 90^\circ$. Then altitude $RP$ is drawn to hypotenuse $SQ$. So we have similar triangles: $\triangle RQP \sim \triangle SRQ \sim \triangle RSP$.

Step2: Analyze $\triangle RQP$

In $\triangle RQP$, which is right-angled at $P$, $\cos Q = \frac{\text{adjacent to } Q}{\text{hypotenuse of } \triangle RQP}$. The adjacent side to $Q$ is $PQ$, and the hypotenuse of $\triangle RQP$ is $RQ$? Wait, no. Wait, in $\triangle SRQ$ (right at $R$), $\cos Q = \frac{RQ}{SQ}$? Wait, no, $\angle Q$ is in $\triangle SRQ$, so adjacent side is $RQ$? Wait, no, hypotenuse is $SQ$, adjacent is $RQ$? Wait, no, in right triangle $SRQ$, right-angled at $R$, so sides: $SR$ and $RQ$ are legs, $SQ$ is hypotenuse. So $\cos Q = \frac{\text{adjacent to } Q}{\text{hypotenuse}} = \frac{RQ}{SQ}$? Wait, no, adjacent to $Q$ is $RQ$? Wait, no, $\angle Q$: the sides: opposite is $SR$, adjacent is $RQ$, hypotenuse is $SQ$. So $\cos Q = \frac{RQ}{SQ}$. But also, in $\triangle RQP$, which is right-angled at $P$, $\cos Q = \frac{PQ}{RQ}$? Wait, let's check similarity.

Since $RP$ is altitude to hypotenuse $SQ$ in right $\triangle SRQ$, then $\triangle RQP \sim \triangle SRQ$ (AA similarity: both right-angled, share $\angle Q$). So corresponding sides are proportional. In $\triangle SRQ$, $\cos Q = \frac{RQ}{SQ}$. In $\triangle RQP$, $\cos Q = \frac{PQ}{RQ}$ (since in $\triangle RQP$, adjacent to $Q$ is $PQ$, hypotenuse is $RQ$). Wait, let's verify.

In $\triangle RQP$ (right at $P$), $\cos Q = \frac{PQ}{RQ}$ (adjacent over hypotenuse). Let's check the options. The options are:

1. $\frac{RP}{RQ}$: that's $\sin Q$ in $\triangle RQP$, since $\sin Q = \frac{RP}{RQ}$.

1. $\frac{SR}{RQ}$: that's $\tan Q$ in $\triangle SRQ$ (since $\tan Q = \frac{SR}{RQ}$).

1. $\frac{PR}{SP}$: that's $\tan \angle S$, maybe.

1. $\frac{PQ}{RQ}$: that's $\cos Q$ in $\triangle RQP$, which is similar to $\triangle SRQ$, so $\cos Q$ should be equal to that.

So the correct ratio is $\frac{PQ}{RQ}$.

Answer:

$\frac{PQ}{RQ}$ (the fourth option: $\boldsymbol{\frac{PQ}{RQ}}$)