Question

given right triangle xye, what is the value of tan(y)? four option boxes, image of right triangle xye with right angle at e, angle at x 60°, angle at y 30°, side xy=4

Explanation:

Step1: Identify triangle properties

In right triangle \( XYE \), \( \angle E = 90^\circ \), \( \angle X = 60^\circ \), so \( \angle Y = 30^\circ \) (since triangle angles sum to \( 180^\circ \)). Hypotenuse \( XY = 4 \).

Step2: Find opposite and adjacent sides to \( \angle Y \)

For \( \angle Y = 30^\circ \), opposite side \( = XE \), adjacent side \( = YE \). In a 30 - 60 - 90 triangle, sides are in ratio \( 1:\sqrt{3}:2 \). Hypotenuse \( XY = 4 \), so opposite side to \( 30^\circ \) ( \( XE \)) is \( \frac{4}{2}=2 \), adjacent side ( \( YE \)) is \( 2\sqrt{3} \)? Wait, no—wait, \( \angle Y = 30^\circ \), so angle at \( Y \): opposite side is \( XE \), adjacent is \( YE \). Wait, maybe better to use angle \( Y = 30^\circ \), so \( \tan(Y)=\tan(30^\circ)=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3} \)? Wait, no, wait the triangle: \( \angle X = 60^\circ \), \( \angle Y = 30^\circ \), right angle at \( E \). So for angle \( Y \), the opposite side is \( XE \), adjacent is \( YE \). Let's find lengths: hypotenuse \( XY = 4 \). So \( \sin(30^\circ)=\frac{XE}{XY}\Rightarrow XE = 4\times\frac{1}{2}=2 \). \( \cos(30^\circ)=\frac{YE}{XY}\Rightarrow YE = 4\times\frac{\sqrt{3}}{2}=2\sqrt{3} \). Then \( \tan(Y)=\frac{opposite}{adjacent}=\frac{XE}{YE}=\frac{2}{2\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3} \)? Wait, but maybe I mixed up angles. Wait, angle at \( X \) is \( 60^\circ \), so angle at \( Y \) is \( 30^\circ \). Wait, no—wait, in triangle \( XYE \), right - angled at \( E \), so \( \angle X+\angle Y = 90^\circ \). So if \( \angle X = 60^\circ \), then \( \angle Y = 30^\circ \). Now, \( \tan(Y)=\tan(30^\circ)=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3} \)? Wait, but maybe the triangle is labeled differently. Wait, maybe \( XY = 4 \), and we can also consider angle at \( Y \) as \( 30^\circ \), so the sides: opposite to \( 30^\circ \) is \( XE \), adjacent is \( YE \). Alternatively, maybe I made a mistake. Wait, let's re - express: in a right triangle, \( \tan(\theta)=\frac{\text{opposite}}{\text{adjacent}} \). For \( \angle Y \), opposite side is \( XE \), adjacent is \( YE \). If \( \angle X = 60^\circ \), then \( \angle Y = 30^\circ \), \( XY = 4 \) (hypotenuse). So \( XE = XY\sin(30^\circ)=4\times\frac{1}{2}=2 \), \( YE = XY\cos(30^\circ)=4\times\frac{\sqrt{3}}{2}=2\sqrt{3} \). Then \( \tan(Y)=\frac{XE}{YE}=\frac{2}{2\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3} \approx 0.577 \), but the options seem to have \( \frac{1}{\sqrt{3}} \) or \( \frac{\sqrt{3}}{3} \) (maybe the options are \( 1 \), \( \frac{\sqrt{3}}{3} \), \( \sqrt{3} \), \( \frac{\sqrt{3}}{2} \)? Wait, the user's options are a bit unclear, but assuming the standard 30 - 60 - 90 triangle, \( \tan(30^\circ)=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3} \). Wait, maybe I mixed up the angle: if angle at \( Y \) is \( 60^\circ \), then angle at \( X \) is \( 30^\circ \). Let's check: if \( \angle Y = 60^\circ \), then \( \tan(60^\circ)=\sqrt{3} \). Wait, the triangle has \( \angle X = 60^\circ \), right angle at \( E \), so \( \angle Y = 30^\circ \). So \( \tan(Y)=\tan(30^\circ)=\frac{\sqrt{3}}{3} \) (or \( \frac{1}{\sqrt{3}} \)).

Answer:

\(\frac{\sqrt{3}}{3}\) (or \(\frac{1}{\sqrt{3}}\) depending on the exact option labels, but the value of \(\tan(30^\circ)\) is \(\frac{\sqrt{3}}{3}\))