# Thread: The Ambiguous Case (law of sine/ cosine)

1. ## The Ambiguous Case (law of sine/ cosine)

need some help 3 probs. on setting up the pictures, etc. plz thxs!

2. Hello Nismo
Originally Posted by Nismo
need some help 3 probs. on setting up the pictures, etc. plz thxs!

For question 1, I am assuming that the ship's speed of 16 mph is measured through the water - not relative to the earth.

The vector law of addition (of velocities) states:
The velocity of the ship relative to the earth $({_SV_E})$ = the velocity of the ship relative to the water $({_SV_W})$ + the velocity of the water relative to the earth $({_WV_E})$.
Study the attached diagram carefully. It shows these three velocity vectors, drawn with the two possible directions of the current - that is, the velocity of the water relative to the earth.

Using the Sine Rule:
$\frac{\sin\theta}{16}=\frac{\sin 15}{14}$
This gives the value of $\theta$ (and $180-\theta$). From this you can work out the two directions the current makes with the North. (If my arithmetic is correct, they are $32^o$ and $178^o$, to the nearest degree.)

In question 2, you need to draw a similar diagram, but this time you'll find there is only only possibility because the triangle is right-angled. See the second of the attachments. Since it's a right-angle triangle, it's very simple to work out the length of the third side to give the ground speed of the ship (that's the speed relative to the earth).

Lastly, look at my third diagram to see the two different possibilities. Find $\theta$ (and $180-\theta$) using the Sine Rule. Then you can either (a) find the third angle and then use the Cosine Rule on each triangle to find the distances along the ground, or (b) possibly easier, because it just uses right-angled triangles, find the height above ground of the top of the pole (the dotted line in my diagram) and then the horizontal distances.

Can you finish all these now?