Question

in the diagram below, $\triangle npm \sim \triangle qrm$. which statement is true?
answer
$\circ \tan r = \frac{mn}{mp}$ $\circ \tan p = \frac{mr}{mq}$
$\circ \cos q = \frac{mp}{mn}$ $\circ \sin n = \frac{mq}{qr}$

Explanation:

Step1: Recall Similar Triangles Properties

Since $\triangle NPM \sim \triangle QRM$, corresponding angles are equal. So $\angle P = \angle R$, $\angle N = \angle Q$, $\angle PMN = \angle RMQ = 90^\circ$.

Step2: Analyze Trigonometric Ratios

• For $\tan R$: In $\triangle QRM$, $\tan R = \frac{MQ}{MR}$. But from similarity, $\frac{MQ}{MR}=\frac{MN}{MP}$ (corresponding sides of similar triangles). Wait, no, let's check each option:
• Option 1: $\tan R = \frac{MN}{MP}$. Since $\triangle NPM \sim \triangle QRM$, $\angle R = \angle P$. In $\triangle NPM$, $\tan P = \frac{MN}{MP}$, so $\tan R = \tan P = \frac{MN}{MP}$. Wait, but let's check other options.
• Option 2: $\tan P = \frac{MR}{MQ}$. In $\triangle NPM$, $\tan P = \frac{MN}{MP}$, and from similarity, $\frac{MN}{MP}=\frac{MQ}{MR}$, so $\tan P = \frac{MQ}{MR}$, not $\frac{MR}{MQ}$. So this is wrong.
• Option 3: $\cos Q = \frac{MP}{MN}$. In $\triangle NPM$, $\cos N = \frac{MN}{PN}$, and $\angle N = \angle Q$, so $\cos Q = \cos N = \frac{MN}{PN}$, not $\frac{MP}{MN}$. Wrong.
• Option 4: $\sin N = \frac{MQ}{QR}$. In $\triangle NPM$, $\sin N = \frac{MP}{PN}$, and from similarity, $\frac{MP}{PN}=\frac{MR}{QR}$, not $\frac{MQ}{QR}$. Wrong.

Wait, rechecking Option 1: Since $\triangle NPM \sim \triangle QRM$, $\angle R = \angle P$. In right triangle $NPM$, $\tan P = \frac{MN}{MP}$ (opposite/adjacent). So $\tan R = \tan P = \frac{MN}{MP}$. So Option 1 is correct? Wait, no, let's re-express.

Wait, $\triangle NPM$: right-angled at M, so $\tan P = \frac{MN}{MP}$. $\triangle QRM$: right-angled at M, so $\tan R = \frac{MQ}{MR}$. But from similarity, $\frac{MN}{MP}=\frac{MQ}{MR}$, so $\tan R = \frac{MQ}{MR} = \frac{MN}{MP}$ (since $\frac{MN}{MP}=\frac{MQ}{MR}$). So $\tan R = \frac{MN}{MP}$, which is Option 1. Wait, but let's check again.

Wait, maybe I made a mistake. Let's list the correct ratios:

In $\triangle NPM$ (right at M):

• $\tan P = \frac{MN}{MP}$
• $\sin N = \frac{MP}{PN}$
• $\cos N = \frac{MN}{PN}$

In $\triangle QRM$ (right at M):

• $\tan R = \frac{MQ}{MR}$
• $\sin Q = \frac{MR}{QR}$
• $\cos Q = \frac{MQ}{QR}$

Since $\triangle NPM \sim \triangle QRM$, $\frac{MN}{MP} = \frac{MQ}{MR}$, $\frac{MP}{PN} = \frac{MR}{QR}$, $\frac{MN}{PN} = \frac{MQ}{QR}$.

Now check each option:

1. $\tan R = \frac{MN}{MP}$: $\tan R = \frac{MQ}{MR}$, and $\frac{MQ}{MR} = \frac{MN}{MP}$ (from similarity), so $\tan R = \frac{MN}{MP}$. Correct.

1. $\tan P = \frac{MR}{MQ}$: $\tan P = \frac{MN}{MP} = \frac{MQ}{MR}$ (from similarity), so $\tan P = \frac{MQ}{MR}$, not $\frac{MR}{MQ}$. Wrong.

1. $\cos Q = \frac{MP}{MN}$: $\cos Q = \cos N = \frac{MN}{PN}$, and $\frac{MN}{PN}

eq \frac{MP}{MN}$. Wrong.

1. $\sin N = \frac{MQ}{QR}$: $\sin N = \frac{MP}{PN} = \frac{MR}{QR}$ (from similarity), so $\sin N = \frac{MR}{QR}$, not $\frac{MQ}{QR}$. Wrong.

So the correct option is $\tan R = \frac{MN}{MP}$.

Answer:

$\tan R = \frac{MN}{MP}$ (the first option: $\boldsymbol{\tan R = \frac{MN}{MP}}$)