Chapter 12.7, Problem 53SR

To determine

To solve the given triangle and round side length to the nearest tenth and angle to the nearest degree.

Answer to Problem 53SR

  $q=11.4$

  $S=42°$

  $R=67°$

Explanation of Solution

Given:

  

Concept used:

Law of cosines to find the missing side length:

  $q^2=s^2+r^2-2rs\cos Q$

Law of sines to find the missing angles:

  $\frac{\sin Q}{q}=\frac{\sin R}{r}=\frac{\sin S}{s}$

Calculation:

From the given triangle,

Let

  $\begin{array}{l}r=21.2\\ s=15.5\end{array}$

  $\angle Q=31°$

Using the law of cosines,

  $\begin{array}{c}{q}^2={15.5}^2+{21.2}^2-2(15.5)(21.2)\cos 31\\ =240.25+449.44-657.2\cos (0.85)\\ =240.25+449.44-657.2(0.85)\\ =240.25+449.44-558.62\\ q^2=131.07\end{array}$

If $q^2=131.07$ then $q=11.44$

Rounding off to the nearest tenth, $q=11.4$

Using the law of sines,

Consider,

  $\begin{array}{c}\frac{\sin Q}{q}=\frac{\sin R}{r}\\ \frac{\sin 31}{11.4}=\frac{\sin R}{21.2}\\ \frac{0.5}{11.4}=\frac{\sin R}{21.2}\\ \frac{0.5\times 21.2}{11.4}=\sin R\\ \frac{10.6}{11.4}=\sin R\\ 0.92=\sin R\\ {\sin }^{-1}0.92=R\\ 66.9=R\\ 67°=R\end{array}$

Consider,

  $\begin{array}{c}\frac{\sin Q}{q}=\frac{\sin S}{s}\\ \frac{\sin 31}{11.4}=\frac{\sin S}{15.5}\\ \frac{0.5}{11.4}=\frac{\sin S}{15.5}\\ \frac{0.5\times 15.5}{11.4}=\sin S\\ \frac{7.75}{11.4}=\sin S\\ 0.67=\sin S\\ {\sin }^{-1}(0.67)=S\\ 42.06=S\\ 42°=S\end{array}$

Conclusion:

Therefore, the missing measures are $q=11.4$ , $S=42°$ and $R=67°.$