Hello, BiGpO6790!

A car is driven east for a distance of 45 km, then north for 29 km,

and then in a direction 29° east of north for 27 km.

Determine:

(a) the magnitude (in km) of the car's total displacement from its starting point

(b) the angle (from east) of the car's total displacement measured from its starting direction. Code:

N * D
: /:
:29°/ :
: /27: y
: / :
:/61° :
C * - - * E
| x :
| :
29 | : 29
| :
A * - - - - - - * - - *
45 B x F

The car started at $\displaystyle A$, drove east 45 km to $\displaystyle B$,

then north 29 km to $\displaystyle C$, then at heading N29°E for 27 km to $\displaystyle D.$

We have: .$\displaystyle AB = 45,\:BC = EF = 29,\:CD = 27$

. . $\displaystyle \angle NCD = 29^o \quad\Rightarrow\quad \angle DCE = 61^o$

Let $\displaystyle x = CE = BF,\:y = DE$

Draw line segment $\displaystyle AD.$

Its length is the displacement in part (a).

. . In right triangle $\displaystyle DFA\!:\;\;AD \:=\:\sqrt{(x+45)^2 + (y+29)^2}$ .[1]

In right triangle $\displaystyle DEC\!:\;\;\begin{Bmatrix}x \:=\:27\cos61^o \:\approx\:13.09 \\ y \:=\:27\sin61^o \:\approx\:23.61\end{Bmatrix}$

Substitute into [1]: /$\displaystyle AD \:=\:\sqrt{(13.09+ 45)^2 + (23.61 + 29)^2} \:=\:\sqrt{58.09^2 + 52.61^2} \:\approx\:78.37\text{ km}$ .(a)

We want: .$\displaystyle \theta = \angle DAF$

$\displaystyle \tan\theta \:=\:\frac{DF}{AF} \:=\:\frac{52.61}{58.09} \:=\:0.905663625$

Therefore: .$\displaystyle \theta \;\approx\;42.17^o $ .(b)