# Trigonometry

• Mar 25th 2009, 02:25 AM
Mr Rayon
Trigonometry
A ship sails on a bearing of S20°W for 14 km, then changes direction and sails for 20 km and drops anchor. Its bearing from the starting points is now N65°W.

a) How far is it from the starting point?

And yes...I have done a diagram on it.

Any help will be appreciated!
• Mar 25th 2009, 06:32 AM
Trigonometry
Hello Mr Rayon
Quote:

Originally Posted by Mr Rayon
A ship sails on a bearing of S20°W for 14 km, then changes direction and sails for 20 km and drops anchor. Its bearing from the starting points is now N65°W.

a) How far is it from the starting point?

Let's call the starting point $A$, the point where the ship changes direction $B$, and its final point $C$. Then we have:

• $AB = 14 = c$
• $BC = 20= a$
• $\angle BAC = 180 - (20+65) = 95^o = A$

1 Use the Sine Rule to calculate $\angle C$:

• $\frac{\sin C}{c} = \frac{\sin A}{a}$

2 Use the angle sum of a triangle $= 180^o$ to calculate $\angle B$.

3 Then use the Cosine Rule to find the distance $AC$:

• $b^2 = a^2 + c^2 -2ac\cos B$

• Mar 25th 2009, 06:48 AM
earboth
Quote:

Originally Posted by Mr Rayon
A ship sails on a bearing of S20°W for 14 km, then changes direction and sails for 20 km and drops anchor. Its bearing from the starting points is now N65°W.

a) How far is it from the starting point?

...

You are dealing with a triangle of which you know two sides and the angle opposite the larger of the two sides.

1. Use the Sine rule to calculate the angle at the top of the triangle (the lightblue one):

$\dfrac{\sin(\tau)}{\sin(95^\circ)} = \dfrac{14}{20}$

2. Calculate the 3rd angle (the lightgreen one).

3. Use Cosine rule to calculate the distance between the starting point S and the place where the ship lies at anchor (the red line)

For your confirmation: I've got d = 13.115 km