Question

23 in the diagram below. \\( \triangle e r m \sim \triangle j t m \\).
which statement is always true?

1. \\( \cos j = \frac { r m } { r e } \\)
2. \\( \cos r = \frac { j m } { j t } \\)
3. \\( \tan t = \frac { r m } { e m } \\)
4. \\( \tan e = \frac { t m } { j m } \\)

Explanation:

Step1: Recall the definition of cosine

The cosine of an angle in a right - triangle is defined as $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$.

Step2: Analyze $\cos J$ in $\triangle JTM$ and $\triangle JRE$

Since $\triangle ERM\sim\triangle JTM$, then $\angle J$ is common to both $\triangle JTM$ and $\triangle JRE$ (where $\triangle JRE$ is a right - triangle with right - angle at $M$). In $\triangle JRE$, $\cos J = \frac{JM}{JR}$. In $\triangle JTM$, $\cos J=\frac{JM}{JT}$. But we know from similar triangles $\triangle ERM\sim\triangle JTM$ that the ratios of corresponding sides are equal.
Let's check each option:

• Option 1: In $\triangle JRE$, $\cos J=\frac{JM}{JR}

eq\frac{RM}{RE}$ (because $JM$ is adjacent to $\angle J$ and $JR$ is the hypotenuse in $\triangle JRE$).

• Option 2: In $\triangle JTM$, $\angle JTM = 90^{\circ}$. In $\triangle ERM$, $\angle ERM$ is not related to $\frac{JM}{JT}$ in terms of cosine. If we consider $\triangle JTM$ and $\triangle ERM$ (similar), for $\angle R$ in $\triangle ERM$ (right - triangle at $M$), $\cos R=\frac{RM}{ER}$. For $\triangle JTM$, we can use the similarity. Since $\triangle ERM\sim\triangle JTM$, $\angle R=\angle JTM$ (corresponding angles of similar triangles). In $\triangle JTM$, $\cos\angle JTM=\frac{TM}{JT}$. But also, since $\triangle JTM\sim\triangle JRE$ (by AA similarity, $\angle J$ is common and $\angle JTM=\angle JME = 90^{\circ}$), and $\triangle ERM\sim\triangle JTM$, we know that $\cos R=\frac{JM}{JT}$ (because in $\triangle JTM$, if we consider the angle equal to $\angle R$ (from similarity), the adjacent side is $JM$ and hypotenuse is $JT$).
• Option 3: $\angle T$ is not an angle in a right - triangle in a standard sense for the given ratio $\frac{RM}{EM}$. The tangent of an angle is defined in a right - triangle as $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. There is no right - triangle where $\angle T$ is an acute angle and the sides are $RM$ and $EM$.
• Option 4: In $\triangle JTM$, $\tan E$ is not related to $\frac{TM}{JM}$. In $\triangle ERM$ (right - triangle at $M$), $\tan E=\frac{RM}{EM}$. In $\triangle JTM$, $\tan J=\frac{TM}{JM}$.

Answer:

1. $\cos R=\frac{JM}{JT}$