# Math Help - Bearing of wind

1. ## 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

2. 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.

3. 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 $A$ to $B$: $AB = 1560$ km.

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

Let $w$ = wind speed (in km/hr).
The wind blows from $C$ to $B$ in 3 hours: $CB \,=\,3w$

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

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

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

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

Therefore, the wind is blowing from a direction
. . with a bearing of $180^o - 72.5^o \:=\:\boxed{108.5^o}$

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 $A$ to $C.$
Due to the wind, it flies to $D$: . $AD = 1125$

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

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

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

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

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

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