Question

in the right triangle shown, m∠y = 30° and xy = 6. how long is yz? choose 1 answer: a 3√3 b 4√3 c 6√3 d 12 e 12√3

Explanation:

Step1: Identify triangle type and trigonometric ratio

This is a right triangle with \( \angle X = 90^\circ \), \( \angle Y = 30^\circ \), and \( XY = 6 \). We need to find \( YZ \) (hypotenuse). In a right triangle, \( \cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}} \). Here, \( \theta = 30^\circ \), adjacent side to \( \angle Y \) is \( XY = 6 \), and hypotenuse is \( YZ = z \). So, \( \cos(30^\circ)=\frac{XY}{YZ} \).

Step2: Recall \( \cos(30^\circ) \) value and solve for \( YZ \)

We know that \( \cos(30^\circ)=\frac{\sqrt{3}}{2} \). Substituting into the formula: \( \frac{\sqrt{3}}{2}=\frac{6}{z} \). Cross - multiplying gives \( z\times\sqrt{3}=2\times6 = 12 \). Then, \( z=\frac{12}{\sqrt{3}} \). Rationalizing the denominator (multiplying numerator and denominator by \( \sqrt{3} \)): \( z=\frac{12\sqrt{3}}{3}=4\sqrt{3} \)? Wait, no, wait. Wait, maybe I made a mistake. Wait, in a 30 - 60 - 90 triangle, the side opposite 30° is half the hypotenuse. Wait, \( \angle Z = 60^\circ \), \( \angle Y = 30^\circ \), \( XY \) is opposite \( \angle Z \)? Wait, no, let's re - examine the triangle. The right angle is at \( X \), so \( XY \) and \( XZ \) are the legs, \( YZ \) is the hypotenuse. \( \angle Y = 30^\circ \), so the side opposite \( \angle Y \) is \( XZ \), and the side adjacent is \( XY \). Wait, maybe using sine. \( \sin(30^\circ)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{XZ}{YZ} \), and \( \cos(30^\circ)=\frac{XY}{YZ} \). Wait, \( XY = 6 \), \( \cos(30^\circ)=\frac{\sqrt{3}}{2}=\frac{6}{YZ} \), so \( YZ=\frac{6}{\cos(30^\circ)}=\frac{6}{\frac{\sqrt{3}}{2}} = 6\times\frac{2}{\sqrt{3}}=\frac{12}{\sqrt{3}} = 4\sqrt{3} \)? No, wait, that's not right. Wait, no, in a 30 - 60 - 90 triangle, the sides are in the ratio \( 1:\sqrt{3}:2 \), where the side opposite 30° is the shortest side (let's say length \( a \)), the side opposite 60° is \( a\sqrt{3} \), and the hypotenuse is \( 2a \). Wait, \( \angle Y = 30^\circ \), so the side opposite \( \angle Y \) is \( XZ \), and the side adjacent is \( XY \). Wait, if \( \angle X = 90^\circ \), then \( XY \) is one leg, \( XZ \) is the other leg, \( YZ \) is hypotenuse. \( \angle Y = 30^\circ \), so \( \sin(30^\circ)=\frac{XZ}{YZ} \), \( \cos(30^\circ)=\frac{XY}{YZ} \). Wait, \( XY = 6 \), \( \cos(30^\circ)=\frac{\sqrt{3}}{2} \), so \( YZ=\frac{XY}{\cos(30^\circ)}=\frac{6}{\frac{\sqrt{3}}{2}}=\frac{12}{\sqrt{3}} = 4\sqrt{3} \)? But wait, another way: in a 30 - 60 - 90 triangle, the hypotenuse is twice the shorter leg. Wait, which is the shorter leg? The shorter leg is opposite the 30° angle. So if \( \angle Y = 30^\circ \), the side opposite \( \angle Y \) is \( XZ \), so \( XZ=\frac{1}{2}YZ \). The other leg \( XY \) is opposite \( \angle Z = 60^\circ \), so \( XY = XZ\sqrt{3} \). Let \( XZ = a \), then \( XY = a\sqrt{3}=6 \), so \( a=\frac{6}{\sqrt{3}} = 2\sqrt{3} \), and \( YZ = 2a = 4\sqrt{3} \). Wait, but the options have 4√3 as option B. Wait, but earlier when I thought of the wrong ratio, I was confused. So the correct answer is 4√3? Wait, no, wait, maybe I mixed up the adjacent and opposite. Wait, let's look at the triangle again. The right angle is at X, so vertices are X (right angle), Y, Z. So angle at Y is 30°, so side XY is adjacent to angle Y, XZ is opposite to angle Y, and YZ is hypotenuse. So \( \cos(\angle Y)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{XY}{YZ} \). So \( \cos(30^\circ)=\frac{6}{YZ} \), \( YZ=\frac{6}{\cos(30^\circ)}=\frac{6}{\frac{\sqrt{3}}{2}} = 4\sqrt{3} \). Yes, that's correct.

Answer:

B. \( 4\sqrt{3} \)