# Math Help - [SOLVED] Using the sine rule.

1. ## [SOLVED] Using the sine rule.

Town B is 6 km, on a bearing of 020°, from town A. Town C is located on a bearing of 055° from town A and on a bearing of 120° from town B. Work out the distance of town C from (a) town A and (b) town B.
Can someone show me how the diagram is suppose to look like and how to solve this . As I don't really understand how to work with bearings.

thanks!

2. Hello, Tweety!

Town B is 6 km, on a bearing of 020°, from town A.
Town C is located on a bearing of 055° from town A and on a bearing of 120° from town B.
Work out the distance of town C from (a) town A and (b) town B.

Bearings are measured clockwise from North.
Code:
                Q
:
:
B o 120°
P        /: *
|       / :60°*
|      /  :     *
|     /20°:       *
|    /    :     70° o C
|20°/     :     *
|  /      : *
| / 30° * :
|/  *     :
A o         R

$\angle PAB \:=\: 20^o\:=\:\angle ABR,\:AB = 6\text{ km}$

$\angle PAC \:=\:50^o \quad\Rightarrow\quad \angle BAC \:=\:30^o$

$\angle QBC \:=\:120^o \quad\Rightarrow\quad \angle CBR \:=\:60^o\quad\Rightarrow\quad \angle ABC \:=\:80^o$

In $\Delta ABC\!:\;\;\frac{AC}{\sin80^o} \:=\:\frac{6}{\sin70^o} \quad\Rightarrow\quad AC \:=\:\frac{6\sin80^o}{\sin70^o} \:\approx\:6.3\text{ km}$

. . . . . . . . $\frac{BC}{\sin30^o} \:=\:\frac{6}{\sin70^o} \quad\Rightarrow\quad BC \:=\:\frac{6\sin30^o}{\sin70^o} \:\approx\:3.2\text{ km}$

3. edit Soroban's excellent graphic so that angle BAC = 35 instead of 30 and angle BCA = 65 instead of 70.

fix that also in the law of sines equations and you'll be fine.

4. Originally Posted by Tweety
Can someone show me how the diagram is suppose to look like and how to solve this . As I don't really understand how to work with bearings.

thanks!
Bearings are the angles measured from North to clockwise direction. Please see attached diagram.