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
to
:
km.
It must fly to
,
km, due to the wind.
Let
= wind speed (in km/hr).
The wind blows from
to
in 3 hours: 
Law of Cosines: . ^2 \;=\;1560^2 + 1500^2 - 2(1560)(1500)\cos9^o \;=\;61218.56602)
Hence: . 
Law of Cosines: . (BC)} \;=\;\frac{1560^2 + 247.5^2 - 1500^2}{2(1560)(247.5)} \;=\;0.951267483)
Hence: . 
Therefore, the wind is blowing from a direction
. . with a bearing of 
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
to 
Due to the wind, it flies to
: . 
The wind blows from
to
: . 
Law of Sines: . 
Hence: . 
. . Then: . 
Law of Sines: . 
Therefore: . 