Hello, acc100jt!

A plane, whose speed in still air is 500 km/h,

flies from a point A to a point B, 1560 km due north of A.

Because of a constant wind, the plane must head in a direction whose bearing is 9°.

Given that the flight takes 3 hours,

(i) Show that the speed of the wind is approximately 82.5 km/h

(ii) Find the bearing of the direction from which the wind is blowing. Code:

B *
| * 3w
| *
| *
| * C
1560 | *
| *
| * 1500
|9°*
| *
|*
A *

The plane intends to fly from $\displaystyle A$ to $\displaystyle B$: $\displaystyle AB = 1560$ km.

It must fly to $\displaystyle C$, $\displaystyle AC = 1500$ km, due to the wind.

Let $\displaystyle w$ = wind speed (in km/hr).

The wind blows from $\displaystyle C$ to $\displaystyle B$ in 3 hours: $\displaystyle CB \,=\,3w$

Law of Cosines: .$\displaystyle (3w)^2 \;=\;1560^2 + 1500^2 - 2(1560)(1500)\cos9^o \;=\;61218.56602$

Hence: .$\displaystyle 3w \:=\:247.423859\quad\Rightarrow\quad w \:=\:82.47461968 \:\approx\:\boxed{82.5\text{ km/hr}}$

Law of Cosines: .$\displaystyle \cos B \;=\;\frac{AB^2 + BC^2 - AC^2}{(AB)(BC)} \;=\;\frac{1560^2 + 247.5^2 - 1500^2}{2(1560)(247.5)} \;=\;0.951267483$

Hence: .$\displaystyle B \:=\:71.513012931 \:\approx\:71.5^o$

Therefore, the wind is blowing from a direction

. . with a bearing of $\displaystyle 180^o - 72.5^o \:=\:\boxed{108.5^o}$

On another occasion the wind, whose speed is now 90 km/h,

is blowing from a direction whose bearing is 120°.

A second plane, whose speed in still air is 375 km/h,

flies from A to a point C, which is due east of A.

Given that this flight also takes 3 hours.

(iii) find the distance AC Code:

D
* :
* * :
1125 * *60°:
* 270 * :
* * :
* 30° *:
A * * * * * * * * * * * * * * * * C

The plane intends to fly from $\displaystyle A$ to $\displaystyle C.$

Due to the wind, it flies to $\displaystyle D$: .$\displaystyle AD = 1125$

The wind blows from $\displaystyle C$ to $\displaystyle D$: .$\displaystyle CD = 270$

Law of Sines: .$\displaystyle \frac{\sin A}{a} \:=\:\frac{\sin C}{c}\quad\Rightarrow\quad \sin A \:=\:\frac{270\sin30^o}{1125} \:=\:0.12$

Hence: .$\displaystyle A \:=\:6.892102579 \:\approx\:6.9^o$

. . Then: .$\displaystyle D \:=\:180^o - 30^o - 6.9^o \:=\:143.1^o$

Law of Sines: .$\displaystyle \frac{d}{\sin D} \:=\:\frac{c}{\sin C} \quad\Rightarrow\quad d \:=\:\frac{1125\sin143.1^o}{\sin30^o} \:=\:1350.631451$

Therefore: .$\displaystyle AC \;\approx\;\boxed{1350.6\text{ km}}$