Chapter 71, Problem 48AR

To determine

(a)

The length of side m.

Answer to Problem 48AR

The length of side AC is $m=8.484\text{inches}$

Explanation of Solution

Given information:

The given figure is

  

Calculation:

Let us consider the triangle ABC, the length of side AC of length m can be found from cosine law. In triangle ABC,

  $\begin{array}{l}A{C}^2=A{B}^2+B{C}^2-2\times AB\times BC\times \cos B\\ A{C}^2={14.400}^2+{13.900}^2-2\times 14.4\times 13.9\times \cos 34°50\text{'}\\ A{C}^2=207.36+193.21-324.589\\ A{C}^2=71.98\\ AC=\sqrt{71.98}=m\\ m=8.484\end{array}$

  $m=8.484\text{inches}$

Conclusion:

Thus, the length of side AC is $m=8.484\text{inches}$

To determine

(b)

The value of $\angle N$.

Answer to Problem 48AR

The value of angle N is $\angle N=69.36°$

Explanation of Solution

Given information:

The given figure is

  

Calculation:

Let us consider the triangle ABC, the value of angle N can be found from sine law. In triangle ABC,

From the calculated value of side AC ( $m=8.484\text{inches}$ ),

  $\begin{array}{l}\frac{\sin A}{BC}=\frac{\sin B}{AC}=\frac{\sin C}{AB}\\ \frac{\sin \angle N}{13.900}=\frac{\sin 34°50\text{'}}{8.484}=\frac{\sin \angle P}{14.4}\end{array}$

  $\begin{array}{l}\frac{\sin \angle N}{13.900}=\frac{\sin 34°50\text{'}}{8.484}\\ \sin \angle N=\frac{\sin 34°50\text{'}}{8.484}\times 13.900\\ \sin \angle N=0.9358\\ \angle N={\sin }^{-1}\left(0.9358\right)\\ \angle N=69.36°\end{array}$

  $\angle N=69.36°$

Conclusion:

Thus, the value of angle N is $\angle N=69.36°$.

To determine

(c)

The value of $\angle P$.

Answer to Problem 48AR

The value of angle P is $\angle P=75°48\text{'}$

Explanation of Solution

Given information:

The given figure is

  

Calculation:

Let us consider the triangle ABC, the value of angle P can be found from sine law. In triangle ABC,

From the calculated value of side AC ( $m=8.484\text{inches}$ ),

  $\begin{array}{l}\frac{\sin A}{BC}=\frac{\sin B}{AC}=\frac{\sin C}{AB}\\ \frac{\sin \angle N}{13.900}=\frac{\sin 34°50\text{'}}{8.484}=\frac{\sin \angle P}{14.4}\end{array}$

  $\begin{array}{l}\frac{\sin 34°50\text{'}}{8.484}=\frac{\sin \angle P}{14.4}\\ \sin \angle P=\frac{\sin 34°50\text{'}}{8.484}\times 14.4\\ \sin \angle P=0.9694\\ \angle P={\sin }^{-1}\left(0.9694\right)\\ \angle P=75.81°\\ \angle P=75°48\text{'}\end{array}$

  $\angle P=75°48\text{'}$

Conclusion:

Thus, the value of angle P is $\angle P=75°48\text{'}$.