Question

solve the right triangle.
write your answers in simplified, rationalized form. do not round.
xz =
yz =
m∠x = °

Explanation:

Step1: Find \( m\angle X \)

In a right triangle, the sum of angles is \( 180^\circ \). Since \( \angle Y = 90^\circ \) and \( \angle Z = 60^\circ \), we have \( m\angle X=180^\circ - 90^\circ - 60^\circ = 30^\circ \).

Step2: Find \( XZ \) (hypotenuse)

In right triangle \( XYZ \), \( XY = 5\sqrt{6} \) is opposite \( \angle Z = 60^\circ \). Using the sine function: \( \sin Z=\frac{XY}{XZ} \). So \( \sin 60^\circ=\frac{5\sqrt{6}}{XZ} \). Since \( \sin 60^\circ = \frac{\sqrt{3}}{2} \), we solve for \( XZ \):

$$ XZ=\frac{5\sqrt{6}}{\sin 60^\circ}=\frac{5\sqrt{6}}{\frac{\sqrt{3}}{2}} = 5\sqrt{6}\times\frac{2}{\sqrt{3}} = 5\times2\times\sqrt{\frac{6}{3}} = 10\sqrt{2} $$

Step3: Find \( YZ \)

Using the cosine function: \( \cos Z=\frac{YZ}{XZ} \). We know \( \cos 60^\circ=\frac{1}{2} \) and \( XZ = 10\sqrt{2} \), so:

$$ YZ = XZ\times\cos 60^\circ=10\sqrt{2}\times\frac{1}{2} = 5\sqrt{2} $$

Or using the tangent function for \( \angle Z \): \( \tan Z=\frac{XY}{YZ} \), \( \tan 60^\circ=\sqrt{3}=\frac{5\sqrt{6}}{YZ} \), so \( YZ=\frac{5\sqrt{6}}{\sqrt{3}} = 5\sqrt{2} \).

Answer:

\( XZ = 10\sqrt{2} \)
\( YZ = 5\sqrt{2} \)
\( m\angle X = 30^\circ \)