Question

solve the right triangle.
write your answers in simplified, rationalized form. do not round.
rs =
qr =
m∠r =°

Explanation:

Step1: Find angle R

In a right triangle, the sum of angles is \(180^\circ\). Given \(\angle Q = 90^\circ\) and \(\angle S = 30^\circ\), so \(\angle R=180^\circ - 90^\circ - 30^\circ = 60^\circ\).

Step2: Find hypotenuse RS

In a 30 - 60 - 90 triangle, the side adjacent to \(30^\circ\) (here \(QS = 5\sqrt{3}\)) is \(\frac{\sqrt{3}}{2}\) times the hypotenuse. Let \(RS = x\). Then \(\cos(30^\circ)=\frac{QS}{RS}\), and \(\cos(30^\circ)=\frac{\sqrt{3}}{2}\). So \(\frac{\sqrt{3}}{2}=\frac{5\sqrt{3}}{x}\). Cross - multiplying gives \(x\sqrt{3}=10\sqrt{3}\), so \(x = 10\). Or we can use the property that in a 30 - 60 - 90 triangle, the side adjacent to \(30^\circ\) is \(\frac{\sqrt{3}}{2}\) of the hypotenuse. If the side adjacent to \(30^\circ\) (QS) is \(5\sqrt{3}\), and \(\cos(30^\circ)=\frac{QS}{RS}\), then \(RS=\frac{QS}{\cos(30^\circ)}=\frac{5\sqrt{3}}{\frac{\sqrt{3}}{2}} = 10\).

Step3: Find side QR

In a 30 - 60 - 90 triangle, the side opposite to \(30^\circ\) (QR) is half of the hypotenuse. Since hypotenuse \(RS = 10\), then \(QR=\frac{1}{2}\times RS=\frac{1}{2}\times10 = 5\). Or using \(\tan(30^\circ)=\frac{QR}{QS}\), \(\tan(30^\circ)=\frac{1}{\sqrt{3}}\), so \(\frac{1}{\sqrt{3}}=\frac{QR}{5\sqrt{3}}\), cross - multiplying gives \(QR = 5\).

Answer:

\(RS = 10\)
\(QR = 5\)
\(m\angle R = 60^\circ\)