Question

the figure below shows a right triangle: image of right triangle with sides labeled p units (vertical leg), r units (horizontal leg), q units (hypotenuse), angles x° (at horizontal leg and hypotenuse) and y° (at vertical leg and hypotenuse) what is r ÷ p equal to?

Explanation:

Step1: Recall trigonometric ratios

In a right triangle, the tangent of an angle is defined as the ratio of the opposite side to the adjacent side. For angle \( y^\circ \), the opposite side is \( r \) and the adjacent side is \( p \). So, \( \tan(y^\circ)=\frac{r}{p} \). Also, for angle \( x^\circ \), the tangent of \( x^\circ \) is \( \tan(x^\circ)=\frac{p}{r} \), but we need \( \frac{r}{p} \), which is \( \tan(y^\circ) \) or \( \cot(x^\circ) \) (since \( \cot\theta = \frac{1}{\tan\theta} \), so \( \cot(x^\circ)=\frac{r}{p} \) as \( \tan(x^\circ)=\frac{p}{r} \)). But more directly, using the definition of tangent for angle \( y \): \( \tan(y)=\frac{\text{opposite}}{\text{adjacent}}=\frac{r}{p} \), and also \( \tan(x)=\frac{p}{r} \), so \( \frac{r}{p}=\frac{1}{\tan(x)}=\cot(x) \), or equivalently \( \tan(y) \). Since the problem is about the ratio \( r\div p=\frac{r}{p} \), from the right triangle, for angle \( y \), opposite side is \( r \), adjacent is \( p \), so \( \frac{r}{p}=\tan(y^\circ) \), and for angle \( x \), opposite is \( p \), adjacent is \( r \), so \( \frac{r}{p}=\cot(x^\circ) \) (or \( \tan(90^\circ - x^\circ) \) since \( x + y=90^\circ \)).

Step2: Identify the ratio

Since \( r\div p=\frac{r}{p} \), and in the right triangle, for angle \( y \), \( \tan(y)=\frac{r}{p} \), and for angle \( x \), \( \cot(x)=\frac{r}{p} \) (because \( \cot(x)=\frac{\text{adjacent to }x}{\text{opposite to }x}=\frac{r}{p} \)). So \( r\div p \) is equal to \( \tan(y^\circ) \) (or \( \cot(x^\circ) \)).

Answer:

\( \tan(y^\circ) \) (or \( \cot(x^\circ) \))