# Thread: Confusing problem ELPP again

1. ## Confusing problem ELPP again

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

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
I understand the question until this part...
The other airport must not be directly north of the home base...

Sorry, I don't understand the problem . . . are there typos?

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 . . . ??

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
Hello,

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

5. Thanks!