A boat sails 6km from a point P on a bearing 075° . It then sails 4km on a bearing of 050°. Calculate : 1. The distance that the boat is from P
2. Its bearing from P

Step-by-step explanation:

It is often convenient to find the resultant distance using the Law of Cosines. For unknown triangle side c opposite angle C, it tells you ...

  c² = a² +b² -2ab·cos(C)

For the given geometry, we have ...

  c² = 6² +4² -2(6)(4)cos(155°) ≈ 95.503

  c ≈ √95.503 ≈ 9.77 . . . km

The boat is about 9.8 km from P.

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The angle with respect to the initial vector can be found using the Law of Sines.

  sin(B)/b = sin(C)/c

  B = arcsin(b/c·sin(C)) = arcsin(4/9.77·sin(155°))

  B ≈ arcsin(0.17298) ≈ 9.96°

Then the bearing from P will be ...

  75° -10° = 65°

The boat's bearing from P is about 65°.

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Additional comment

It can be convenient to use a calculator capable of adding vectors in (magnitude∠direction) form. Many scientific or graphing calculators can do that. They generally make the assumption that angles are measured counterclockwise from the +x axis.

Bearings are measured clockwise from North. Typically, a map is oriented so that North is up, which means that (x, y) coordinates on a map correspond to (East, North) coordinates.

If bearing angles are used directly in vector calculations, the resulting (x, y) values actually correspond to (North, East) coordinates on a map. The angles still correspond to bearing angles, but the distances North or East need to take this into account. This is why we have labeled the axes the way we have in the attachment.

where is the diagram? shown below where? ?

step-by-step explanation:

114 plus 113 times 2 and 114 times 3