1. ## Trig Question about Bearings

Jason is on a 50m high bridge and sees two boats anchored below. From his position, Boat A has a bearing of 230 degrees and boat B has a bearing of 120 degrees. Jason estimates the angles of depression to be 38 degrees for boat A and 35 degrees for Boat B. How far apart are the boats. Nearest meter.

Im just a bit confused about the bearings part.

Help is appreciated Thanks in advance

2. Hello, ferken!

This is a 3-D trig problem, but we can work in two dimensions at a time.

Jason is on a 50m high bridge and sees two boats anchored below.
From his position, boat A has a bearing of 230° and boat B has a bearing of 120°.
Jason estimates the angles of depression to be 38° for boat A and 35° for Boat B.
How far apart are the boats (nearest meter)?
Jason looks at boat A.
Code:
      W - - - - - * J
38° * |
*   |
*     | 50
*       |
* 38°     |
A * - - - - - * C
Jason is at $\displaystyle J.$
The bridge is: $\displaystyle JC = 50$
$\displaystyle \angle WJA = 38^o = \angle JAC$

In right triangle $\displaystyle JCA\!:\;\;\tan38^o \:=\:\frac{50}{AC} \quad\Rightarrow\quad AC \:=\:\frac{50}{\tan38^o} \:\approx\:64.0$

Jason looks at boat B.
Code:
    J * - - - - E
| * 35°
|   *
50 |     *
|       *
|     35° *
C * - - - - - * B
Jason is at $\displaystyle J.$
The bridge is: $\displaystyle JC = 50$
$\displaystyle \angle EJB = 35^o = \angle JBC$

In right triangle $\displaystyle JCB\!:\;\;\tan35^o \:=\:\frac{50}{BC} \quad\Rightarrow\quad BC \:=\:\frac{50}{\tan35^o} \:\approx\:71.4$

Looking down from the sky, this is the diagram:
Code:
                N
|
|
|
C* 120°
64 *   *
*  110° *  71.4
A*           *
*       *
*   *
*B

Law of Cosines: .$\displaystyle AB^2 \;=\;64^2 + 71.4^2 - 2(64)(71.4)\cos110^o \;=\;12319.75049$

Therefore: .$\displaystyle AB \;=\;110.9943715 \;\approx\;111\text{ m}$