Question

find r.

write your answer as an integer or as a decimal rounded to the nearest tenth.

Explanation:

Step1: Find the measure of angle \( S \)

The sum of angles in a triangle is \( 180^{\circ} \). So, \( \angle S=180^{\circ}-(17^{\circ} + 150^{\circ})=13^{\circ} \) (since \( \angle Q = 180^{\circ}- 30^{\circ}=150^{\circ}\), wait no, no, wait, the triangle angle sum: \(\angle Q = 180^{\circ}-(20^{\circ}+17^{\circ}) = 143^{\circ}\)? No, wait, no, the Law of Sines is used here. Wait, the Law of Sines formula is \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). In \(\triangle SQR\), by the Law of Sines, \(\frac{r}{\sin R}=\frac{QR}{\sin S}\). First, \(\angle S = 180^{\circ}-\angle Q-\angle R\). But \(\angle Q\) is not given directly. Wait, no, the Law of Sines: \(\frac{r}{\sin17^{\circ}}=\frac{7}{\sin20^{\circ}}\) (because in the Law of Sines formula \(\frac{r}{\sin R}=\frac{QR}{\sin\angle QSR}\), where \(R = 17^{\circ}\), \(QR = 7\), \(\angle QSR=20^{\circ}\))

Step2: Solve for \( r \)

From \(\frac{r}{\sin17^{\circ}}=\frac{7}{\sin20^{\circ}}\), we can cross - multiply to get \(r=\frac{7\times\sin17^{\circ}}{\sin20^{\circ}}\). We know that \(\sin17^{\circ}\approx0.292\), \(\sin20^{\circ}\approx0.342\). Then \(r=\frac{7\times0.292}{0.342}=\frac{2.044}{0.342}\approx6.0\)

Answer:

\(r = 6.0\)