# Bearing of wind

• Sep 21st 2007, 05:59 PM
acc100jt
Bearing of wind
A plane, whose speed in still air is 500 km/h, flies from a point A to a point B, 1560km due north of A. Because of the action of a constant wind, the plane must head in a direction whose bearing is 009 degrees. Given that the flight takes 3 hours,
(i) Show that the speed of the wind is approximately 82.5km/h
(ii) Find the bearing of the direction from which the wind is blowing.

On another occasion the wind, whose speed is now 90km/h, is blowing from a direction whose bearing is 120 degrees. 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

Part (i) is easy, but I have difficulty solving part (ii), can anyone help?
Also, for part (iii), the bearing of the wind is 120 degrees, i can't draw the diagram, where is the angle 120 degrees??bearing of vector?
I understand bearing of a fixed point from another fixed point, but bearing of a vector, i'm confused

Thanks
• Sep 21st 2007, 10:20 PM
ticbol
Are there no typos in your posted question?
Bearing 009 degrees? What is that? Is that the same as bearing 9 degrees?

Bearing 9 degrees is North 9deg East.
Bearing is measured from the North axis, going clockwise.
• Sep 22nd 2007, 04:30 AM
Soroban
Hello, acc100jt!

Quote:

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 $\displaystyle 180^o - 72.5^o \:=\:\boxed{108.5^o}$

Quote:

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