1. ## bearings

An attendant in a lighthouse receives a request for aid from a stalled aircraft located 15 miles due east of the lighthouse. The attendant contacts a second boat located 14miles from the lighthouse on a bearing of N23degreesW. What is the distance and the bearing of the rescue boat from the stalled craft?

90-23=67 degrees

a_____________b
15miles east

14 miles(hyp)

2. Hello, cocoknny!

An attendant in a lighthouse receives a request for aid
from a stalled aircraft located 15 miles due east of the lighthouse.
The attendant contacts a second boat located 14 miles from the lighthouse
on a bearing of N23°W.
What is the distance and the bearing of the rescue boat from the stalled craft?
Code:
    B *
*    *
*         *
*    :         *
*23°:               *
14 *  :                     *                   :
* :                             *           :
*:                                   *   θ :
* - - - - - - - - - - - - - - - - - - - - *
L                 15                      A

The lighthouse is at $L$. .The stalled aircraft is at $A.;\;LA = 15$

The rescue boat is at $B.\;\;BL = 14,\;\angle BLA = 113^o$

We want the distance $AB$ and the bearing $\angle\theta.$

Law of Cosines: . $AB^2
;=\;14^2 + 15^2 - 2(14)(15)\cos113^o \;=\;585.107074$

. . Hence: . $\boxed{AB \:\approx\:24.2\text{ miles}}$

Law of Sines: . $\frac{\sin A}{14} \:=\:\frac{\sin113^o}{24.2}\quad\Rightarrow\quad\s in A \:=\:0.532523469$

Hence: . $\angle A \:\approx\:32.2^o\quad\Rightarrow\quad\boxed{\angl e\theta \:=\:57.8^o}$

Therefore, $B$ is $24.2$ miles from $A$ at a bearing of $N57.8^oW$

3. Originally Posted by cocoknny
An attendant in a lighthouse receives a request for aid from a stalled aircraft located 15 miles due east of the lighthouse. The attendant contacts a second boat located 14miles from the lighthouse on a bearing of N23degreesW. What is the distance and the bearing of the rescue boat from the stalled craft?

90-23=67 degrees

a_____________b
15miles east

14 miles(hyp)
Hi,

1. make a sketch (see attachment)

2. use cosine rule to calculate the distance between the 2 crafts:

$d=\sqrt{15^2+14^2-2\cdot 14 \cdot 15 \cdot \cos(113^\circ)}$

3. use sine rule to calculate the bearing (\alpha is the angle between the bearing to the rescue boat and the bearing from the rescue boat to the stalled craft):

$\frac{\sin(\alpha)}{\sin(113^\circ)}=\frac{15}{d}$

Calculkate $\alpha$. You get the bearing by calculating:

$180^\circ - 23^\circ - \alpha$

(I use N = 0°, E = 90°, S = 180°, W = 270°)

You should get b = 122.2° that means E32.2°S

4. ## hello

thanks so much hun, i was gettin lil confused before

5. ## hey

lol, i didn't even do the cosine rule yet. i only did Pythagoras and fundamental trig identities. I guess i'll look in my textbook. Thanks for the assistance.