Question

solve the right triangle.
write your answers in simplified, rationalized form. do not round.
cd=
ce=
m∠o=

Explanation:

Step1: Find \(m\angle D\)

In a right - triangle \(C = 90^{\circ}\), \(E=45^{\circ}\). Using the angle - sum property of a triangle (\(m\angle C+m\angle D + m\angle E=180^{\circ}\)).
\(90^{\circ}+m\angle D + 45^{\circ}=180^{\circ}\), so \(m\angle D=180^{\circ}-(90^{\circ} + 45^{\circ})=45^{\circ}\).

Step2: Use trigonometric ratios to find \(CD\) and \(CE\)

Since \(\sin E=\frac{CD}{DE}\), and \(DE = \sqrt{6}\), \(E = 45^{\circ}\), \(\sin45^{\circ}=\frac{\sqrt{2}}{2}\).
\(\frac{\sqrt{2}}{2}=\frac{CD}{\sqrt{6}}\), then \(CD=\frac{\sqrt{2}\times\sqrt{6}}{2}=\frac{\sqrt{12}}{2}=\frac{2\sqrt{3}}{2}=\sqrt{3}\).
Also, since \(\cos E=\frac{CE}{DE}\), \(\cos45^{\circ}=\frac{\sqrt{2}}{2}\).
\(\frac{\sqrt{2}}{2}=\frac{CE}{\sqrt{6}}\), then \(CE=\frac{\sqrt{2}\times\sqrt{6}}{2}=\frac{\sqrt{12}}{2}=\sqrt{3}\).

Answer:

\(CD=\sqrt{3}\), \(CE = \sqrt{3}\), \(m\angle D=45^{\circ}\)