Chapter 71, Problem 53AR

To determine

The angle $\angle x$.

Answer to Problem 53AR

The angle is $114°12^{\prime }$.

Explanation of Solution

Given:

Angle shown in below figure:

Calculation:

Apply cos rule in triangle ABC as follows:

  $\begin{array}{c}\cos A=\frac{b^2+c^2-a^2}{2bc}\\ \cos 28°=\frac{{\left(5.2\right)}^2+{\left(8.3\right)}^2-a^2}{2\times 5.2\times 8.3}\\ a=4.44^{″}\end{array}$

So, the side $BC=4.44^{″}$.

Apply sine rule intriangle ABC as follows:

  $\begin{array}{l}\frac{\sin A}{BC}=\frac{\sin B}{AC}\\ \frac{\sin 28°}{4.44^{″}}=\frac{\sin B}{5.2}\\ \sin B=0.5498\\ B=33.355°\end{array}$

Angle $\angle BCA$ is calculated as follows:

  $\begin{array}{l}\angle BCA+33.355°+28°=180°\\ \angle BCA=118.6445°\end{array}$

Angle $\angle A^{\prime }BC$ is calculated as follows:

  $\begin{array}{l}\angle A^{\prime }BC+33.355°=93°\\ \angle A^{\prime }BC=59.645°\end{array}$

Apply cos rule in triangle $A^{\prime }BC$ as follows:

  $\begin{array}{c}\cos B=\frac{a^2+c^2-b^2}{2ac}\\ \cos 59.645°=\frac{{\left(4.44\right)}^2+{\left(9.1\right)}^2-{\left(A^{\prime }C\right)}^2}{2\times 4.44\times 9.1}\\ A^{\prime }C=7.85409^{″}\end{array}$

Apply sine rule intriangle $A^{\prime }BC$ as follows:

  $\begin{array}{c}\frac{\sin A^{\prime }}{BC}=\frac{\sin B}{A^{\prime }C}\\ \frac{\sin A^{\prime }}{4.44}=\frac{\sin 59.645°}{7.85409^{″}}\\ A^{\prime }=29.1968°\end{array}$

Angle $\angle BC{A}^{\prime }$ is calculated as follows:

  $\begin{array}{l}\angle BC{A}^{\prime }+59.645°+29.1968°=180°\\ \angle BC{A}^{\prime }=91.158°\end{array}$

Angle $\angle A^{\prime }CD$ is calculated as follows:

  $\begin{array}{l}\angle A^{\prime }CD+91.158°+118.6445°=260°\\ \angle A^{\prime }CD=50.1975°\end{array}$

Apply sine rule intriangle $A^{\prime }CD$ as follows:

  $\begin{array}{c}\frac{\sin D}{A^{\prime }C}=\frac{\sin C}{A^{\prime }D}\\ \frac{\sin D}{7.85409}=\frac{\sin 50.1975°}{8}\\ D=48.9593°\end{array}$

Angle $\angle D{A}^{\prime }C$ is calculated as follows:

  $\begin{array}{l}\angle D{A}^{\prime }C+48.9593°+50.1975°=180°\\ \angle D{A}^{\prime }C=80.843°\end{array}$

Angle $\angle x$ is calculated as follows:

  $\begin{array}{l}\angle x=\angle D{A}^{\prime }C+\angle B{A}^{\prime }C\\ \angle x=80.843°+33.355°\\ \angle x=114.198°\\ \angle x=114°12^{\prime }\end{array}$

Thus, the angle is $114°12^{\prime }$.

Conclusion:

The angle is $114°12^{\prime }$.