Chapter 6.1, Problem 49E

a)

To determine

To find: The distance between Port and the buoy.

Answer to Problem 49E

The distance between Port and the buoy is $1.66345mm$ .

Explanation of Solution

Given information: $A=22°\&B=36°$

Calculation: Since the sum of the internal angle of a triangle is $180°$ .

Hence,

  $\begin{array}{l}m\angle \left(AbuoyB\right)=180°-22°-36°\\ \therefore m\angle \left(AbuoyB\right)=122°\end{array}$

Now, get the distance between port A and the buoy,

  $\begin{array}{l}b=\frac{2.4\times \sin \left(36°\right)}{\sin \left(122°\right)}\\ b=1.66345mm\end{array}$

Therefore, the required distance is $1.66345mm$ .

b)

To determine

To find: The total time taken by the boat form the angle $A=22°\&B=36°$ .

Answer to Problem 49E

The total time taken by the boat is $1\text{hour}11\min$ .

Explanation of Solution

Given information: The boat speed 2 knots from port A to the buoy and 3 knots from the buoy to port B. And $A=22°\&B=36°$

Formula used: $\text{T=}\frac{\text{distance}}{\text{time}}$

Calculation: Get the distance a,

  $\begin{array}{l}a=\frac{2.4\times \sin \left(22°\right)}{\sin \left(122°\right)}\\ a=1.06mm\end{array}$

Hence, the time taken by the boat to travel from port A to the buoy,

  $\begin{array}{l}\frac{1.06mm}{2mm/hr}=0.83hour\\ =49\min 48\sec \end{array}$

Now, get the time take by the boat to travel from the buoy to the port B.

  $\begin{array}{l}\frac{1.06mm}{3mm/hr}=0.35hour\\ =21\min 12\sec \end{array}$

Hence, the total time taken by the boat:

  $\begin{array}{l}=21\min 12\sec +49\min 48\sec \\ =1\text{hour}11\min \end{array}$

c)

To determine

To compare: The distance between $d_1\&d_2$ .

Answer to Problem 49E

  $d_2$ is $0.1mm$ greater than $d_1$ .

Explanation of Solution

Given information: For $d_1$ , $A=24°\&B=38°$ and for $d_2$ , $A=28°\&B=40°$

Calculation: Get the $d_1$ when $A=24°\&B=38°$

  $m\angle {\left(AbuoyB\right)}_1=118°$

Now, get the $b_1$ ,

  $\begin{array}{l}{b}_1=\frac{2.4\times \sin \left(38°\right)}{\sin \left(118°\right)}\\ b_1=1.673mm\end{array}$

Hence, $d_1$ ,

  $\begin{array}{l}{d}_1=b_1\sin A\\ =1.673\sin \left(24°\right)\\ d_1=0.68mm\end{array}$

Similarly, get the $d_2$ , $A=28°\&B=40°$

  $m\angle {\left(AbuoyB\right)}_1=112°$

  $\begin{array}{l}{b}_2=\frac{2.4\times \sin \left(40°\right)}{\sin \left(112°\right)}\\ b_2=1.664mm\end{array}$

Hence, $d_2$ ,

  $\begin{array}{l}{d}_2=b_2\sin A\\ =1.664\sin \left(28°\right)\\ d_2=0.78mm\end{array}$

Therefore, $d_2$ is $0.1mm$ greater than $d_1$ .