Printable View
A reconnaissance airplane leaves its airport on the east coast of the United States and flies in a direction of 85 degrees. Because of bad weather it returns to another airport 230km to the north of its home base. For the return it flies in a direction of 283 degrees. What is the total distance flown?
I dont know the diagram ... & i dunt undersand T_T
I understand the question until this part...Quote:
Originally Posted by ^_^Engineer_Adam^_^
The other airport must not be directly north of the home base...
Hello, ^_^Engineer_Adam^_^!
Sorry, I don't understand the problem . . . are there typos?
Quote:
A reconnaissance airplane leaves its airport and flies in a direction of 85°.
Because of bad weather, it returns to another airport 230km to the north of its home base.
For the return, it flies in a direction of 283 degrees.
What is the total distance flown?
I assume that a direction of 85° is a bearing, measured CW from North.
So the plane flies from $\displaystyle A$ to $\displaystyle B$.
Code::
:
: o B
: *
: 85° *
A o
Then it flies to airport $\displaystyle C$, 230 km to the north of $\displaystyle A$ (but not directly north).Code:C
o
* *
* *
* *
230 * *
* *
* *
* o B
* *
* *
A o
The return is $\displaystyle CA$ . . . in a southwesterly direction.
But we are told that the return is at a heading of 285°,
. . which is to the northwest . . . ??
Hello,Quote:
Originally Posted by ^_^Engineer_Adam^_^
I've attached a diagram to show you what I've calcualted.
With your triangle ABC you know the length of AC = 230 km. You can calculate from the given bearings the angles in the triangle: 85°, 18° and 77°.
Now use the Sine rule.
$\displaystyle \frac{AB}{230 km}=\frac{\sin(77^\circ)}{\sin(18^\circ)}$
$\displaystyle \frac{BC}{230 km}=\frac{\sin(85^\circ)}{\sin(18^\circ)}$
Solve for AB and BC.
The total length flown by the plane ist AB + BC. I got 1466.7 km.
Greetings
EB
Thanks! :D :D :D :D :D :D :D
Smiley abuse ! :mad:Quote:
Originally Posted by ^_^Engineer_Adam^_^
-=USER WARNED=-
