Two radar stations A and B are 16 km apart and A is due north to B. A ship is know to be on a bearing of 150˚ from A and 10 km from B. Show that this information gives two positions for the ship, and calculate the distance between these two positions.

2. Originally Posted by geton
Two radar stations A and B are 16 km apart and A is due north to B. A ship is know to be on a bearing of 150˚ from A and 10 km from B. Show that this information gives two positions for the ship, and calculate the distance between these two positions.

Hi geton!

Take a look at the diagram. As the ship is 150˚ from A, we can see that this is 30˚ off a 180˚ rotation, so we have angle $\angle BAS =30^{\circ}$.

Now we have a few values, but we need more angles before we can solve for AS.

Angle ASB (by the sine rule):

$\frac{\sin{30^{\circ}}}{10}=\frac{\sin{\angle ASB}}{16}$

$\sin{\angle ASB}=\frac{\frac{1}{2}}{10}\times 16$

$\sin{\angle ASB} = \frac{4}{5}$

$\angle ASB = \sin^{-1}{\left(\frac{4}{5}\right)}$

There are many values of $\sin^{-1}{\left(\frac{4}{5}\right)}$, the reference angle for which is $53.13^{\circ}$. To be exact, infinite, because the function is periodic. The solutions we want are:

$\angle ASB=53.13^{\circ}$ or $\angle ASB=180-53.13=126.87$

Because those are the quadrants where the sine function is positive and are within the reasonable limits for the angles of a triangle.

Case 1: $\angle ASB = 53.13^{\circ}$

Then $\angle ABS = 180-53.13-30=96.87^{\circ}$

Then by the sine rule again,

$\frac{AS}{\sin{96.87^{\circ}}}=\frac{10}{\sin{30}}$

$AS=19.86$km.

Case 2: $\angle ASB = 126.87^{\circ}$

Then $\angle ABS = 180-126.87-30=23.13^{\circ}$

$\frac{AS}{\sin{23.13^{\circ}}}=\frac{10}{\sin{30}}$

$AS=7.86$km.

So the difference between the answers is 12 km. I think.

3. Thank you so much (DivideBy0).