Chapter 70, Problem 47A

To determine

The angle $\angle x$.

Answer to Problem 47A

The angle is $70°38^{\prime }$.

Explanation of Solution

Given:

Angle shown in below figure:

Calculation:

Apply cos rule in triangle ABC as follows:

  $\begin{array}{c}\cos A=\frac{b^2+c^2-a^2}{2bc}\\ \cos 32°10^{\prime }=\frac{{\left(4.2\right)}^2+{\left(6.6\right)}^2-a^2}{2\times 4.2\times 6.6}\\ a=3.777^{″}\end{array}$

So, the side $BC=3.777^{″}$.

Apply sine rule intriangle ABC as follows:

  $\begin{array}{l}\frac{\sin A}{BC}=\frac{\sin B}{AC}\\ \frac{\sin 32°10^{\prime }}{3.777}=\frac{\sin B}{4.2}\\ \sin B=0.5920075\\ B=36.299°\end{array}$

Angle $\angle BCA$ is calculated as follows:

  $\begin{array}{l}\angle BCA+36.299°+32°10^{\prime }=180°\\ \angle BCA=111.53433°\end{array}$

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

  $\begin{array}{l}\angle A^{\prime }BC+36.299°=102°\\ \angle A^{\prime }BC=65.701°\end{array}$

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

  $\begin{array}{c}\cos B=\frac{a^2+c^2-b^2}{2ac}\\ \cos 65.701°=\frac{{\left(3.777\right)}^2+{\left(7.5\right)}^2-{\left(A^{\prime }C\right)}^2}{2\times 3.777\times 7.5}\\ A^{\prime }C=6.87039^{″}\end{array}$

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

  $\begin{array}{c}\frac{\sin A^{\prime }}{BC}=\frac{\sin B}{A^{\prime }C}\\ \frac{\sin A^{\prime }}{3.777}=\frac{\sin 65.701°}{6.87039}\\ A^{\prime }=30.069°\end{array}$

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

  $\begin{array}{l}\angle BC{A}^{\prime }+65.701°+30.069°=180°\\ \angle BC{A}^{\prime }=84.229°\end{array}$

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

  $\begin{array}{l}\angle A^{\prime }CD+84.229°+111.53433°=260°\\ \angle A^{\prime }CD=64.236°\end{array}$

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

  $\begin{array}{c}\frac{\sin D}{A^{\prime }C}=\frac{\sin C}{A^{\prime }D}\\ \frac{\sin D}{6.87039}=\frac{\sin 64.236°}{6.4}\\ D=75.1912°\end{array}$

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

  $\begin{array}{l}\angle D{A}^{\prime }C+75.1912°+64.236°=180°\\ \angle D{A}^{\prime }C=40.5727°\end{array}$

Angle $\angle x$ is calculated as follows:

  $\begin{array}{l}\angle x=\angle D{A}^{\prime }C+\angle B{A}^{\prime }C\\ \angle x=40.5727°+30.069°\\ \angle x=70.64179°\\ \angle x=70°38^{\prime }\end{array}$

Thus, the angle is $70°38^{\prime }$.

Conclusion:

The angle is $70°38^{\prime }$.