Question

solve the right triangle.

write your answers in simplified, rationalized form. do not round.

de =

cd =

m∠e = °

Explanation:

Step1: Find \(m\angle E\)

In a right - triangle, the sum of the two non - right angles is \(90^{\circ}\).
If \(\angle C = 30^{\circ}\), then \(m\angle E=90^{\circ}-\angle C\)
\(m\angle E = 90^{\circ}-30^{\circ}=60^{\circ}\)

Step2: Find \(DE\)

We know that \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). In \(\triangle CDE\) with \(\theta = 30^{\circ}\) and hypotenuse \(CE = 4\sqrt{17}\)
\(\sin30^{\circ}=\frac{DE}{CE}\)
Since \(\sin30^{\circ}=\frac{1}{2}\), then \(DE=\frac{1}{2}\times CE\)
Substitute \(CE = 4\sqrt{17}\) into the formula: \(DE=\frac{1}{2}\times4\sqrt{17}=2\sqrt{17}\)

Step3: Find \(CD\)

We know that \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). In \(\triangle CDE\) with \(\theta = 30^{\circ}\) and hypotenuse \(CE = 4\sqrt{17}\)
\(\cos30^{\circ}=\frac{CD}{CE}\)
Since \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\), then \(CD=\frac{\sqrt{3}}{2}\times CE\)
Substitute \(CE = 4\sqrt{17}\) into the formula: \(CD=\frac{\sqrt{3}}{2}\times4\sqrt{17}=2\sqrt{51}\)

Answer:

\(DE = 2\sqrt{17}\)
\(CD=2\sqrt{51}\)
\(m\angle E = 60^{\circ}\)