# Thread: [SOLVED] Triangles and bearing

1. ## [SOLVED] Triangles and bearing

A ship rounds a headland by sailing first four nautical miles on a course of 069° then five nautical miles on a course of 295°.
Calculate the distance and bearing of its new position from its original position.

2. Originally Posted by tishtash7
A ship rounds a headland by sailing first four nautical miles on a course of 069° then five nautical miles on a course of 295°.
Calculate the distance and bearing of its new position from its original position.
Bearings are drawn clockwise with respect to North. See attachment.

1. Calculate the value of the green angle. (295° - 180° - 69°)

2. Use the Cosine rule to calculate the distance d. (Thick black line)

3. Use Sine rule to calculate the final bearing (red sector).

3. Hello, tishtash7!

Did you make a sketch?

A ship rounds a headland by sailing first four nautical miles on a course of 069°
then five nautical miles on a course of 295°.
Calculate the distance and bearing of its new position from its original position.
Code:
    C *                   Q
*  *               :
*     *           :
*        *    65°:
*     P     *   :
*    :     46° * B
*   :       * :
*  :     *   :
* :69°*  69°:
*: *       :
*         :
A         R

The ship sails 4 miles from $A$ to $B.$
. . $AB \,=\,4,\;\angle PAB \,=\, 69^o \,=\,\angle ABR$

Then it sails 5 miles from $B$ to $C.$
. . $BC \,=\,5,\;\angle QBC \:=\:360^o - 295^o \:=\:65^o$

Hence: . $\angle CBA \:=\:180^o - 69^o - 65^o \:=\:46^o$

In $\Delta ABC$ use the Law of Cosinse:

. . $AC^2 \:=\:4^2 + 5^2 - 2(4)(5)\cos46^o \:=\:13.21366518$

Therefore:. $AC \:\approx\:\boxed{3.6\text{ miles}}$

Law of Cosines again: . $\cos(\angle CAB) \:=\:\frac{3.6^2 + 4^2-5^2}{2(3.6)(4)} \:=\:0.1375$
. . Hence: . $\angle CAB \:\approx\:82^o$

Then: . $\angle CAP \:=\:82^o - 69^o \:=\:13^o$

Therefore, the bearing is: . $360^o - 13^o \:=\:\boxed{347^o}$

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