Math 1432 Homework Help

First Unit Material

Question 4 Section 2.5

The bearing from City A to City B is N$38^{\circ}$E. The bearing from City B to City C is S$52^{\circ}$E. An automobile driven at 65 miles per hour takes 1.4 hours to drive from City A to City B and takes 1.8 hours to drive from City B to City C. Find the distance from City A to City C.

Solution

Remember distance equals the product of speed and time. $$d=rt$$

Therefore, the distance from City A to City B is: $$d_1=m(A,B)=(65)(1.4)=91$$ and the distance from City B to City C is: $$d_2=m(B,C)=(65)(1.8)=117$$

In the drawing above the angle for B can be found by adding the two angles: $$38+52=90$$ This means triangle(A,B,C) is a right triangle. The legs of the right triangle is $d_1$ and $d_2$. The hypotenuse, $r$, is the distance from City A to City C. Therefore, we want to solve for $r$:

$$\begin{align} (d_1)^2+(d_2)^2 & = r^2\\ r^2 & = 91^2+117^2\\ r & = \pm \sqrt{91^2+117^2}\\ & = \pm 148.2228052628879 \end{align}$$

We will use the positive distance.

Answer

Since the answer should be to the nearest mile the answer is: "The distance from City A to City C is approximately 148 miles."

d14:65*1.4;
d24:65*1.8;
38+52;
float(solve((d14)^2+(d24)^2=r^2,r));

\[\tag{${\it \%o}_{41}$}91.0\]

\[\tag{${\it \%o}_{42}$}117.0\]

\[\tag{${\it \%o}_{43}$}90\]

rat: replaced 21970.0 by 21970/1 = 21970.0

\[\tag{${\it \%o}_{44}$}\left[ r=-148.2228052628879 , r=148.2228052628879 \right] \]

Question 5 Section 2.5

Two ships leave a port at the same time. The first ship sails on a bearing of $40^{\circ}$ at 12 knots (nautical miles per hour) and the second on a bearing of $130^{\circ}$ at 14 knots. How far apart are they after 1.5 hours?

Solution

Remember distance equals the product of speed and time. $$d=rt$$ Therefore, the distance for the first boat is $$d_1=(12)(1.5)=18$$ and the distance for the second boad is $$d_2=(14)(1.5)=21$$ These are both in units of nautical miles.

Two rays can be created: ray(Port,Ship 1) and ray(Port, Ship 2). The angles between the two rays is: $$130-40=90$$

This means triangle(Port, Ship 1, Ship 2) is a right triangle. The distance for each leg is $d_1$ and $d_2$. The hypotenuse, $r$, is the distance away from each ship. Therefore, we want to solve for $r$:

$$\begin{align} (d_1)^2+(d_2)^2 & = r^2\\ r^2 & = 18^2+21^2\\ r & = \pm \sqrt{18^2+21^2}\\ & = \pm 27.65863337187866 \end{align}$$

We will use the positive distance.

Answer

Since the answer should be to the nearest nautical mile the answer is: "After 1.5 hours, the ships are 28 nautical miles apart."

d15:12*1.5;
d25:14*1.5;
130-40;
float(solve((d15)^2+(d25)^2=r^2,r));

\[\tag{${\it \%o}_{45}$}18.0\]

\[\tag{${\it \%o}_{46}$}21.0\]

\[\tag{${\it \%o}_{47}$}90\]

rat: replaced 765.0 by 765/1 = 765.0

\[\tag{${\it \%o}_{48}$}\left[ r=-27.65863337187866 , r=27.65863337187866 \right] \]

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