trigonometry question (line of sight, bearing, ect..)?

Question

"the navigator of a ship on a 44 degree course sights a buoy with bearing 134 degrees. after the ship sails 15 km along the same course, the navigator sights the same buoy bearing 168 degrees. find the distance between the ship and the buoy at the time of each sighting".
I have no idea how to solve this, so please explain thoroughly

Answer 1

Bearing is measured from North (corresponding to the y-axis on the Cartesian plane) in a clockwise direction. So the ship is traveling in a NE direction at an angle of 44 degrees from the true North. At the first sighting, the line from the ship to the buoy is one leg of a triangle. Let it be x km long. The second leg of the triangle is the direction of the boat and will be 15 km long when the boat reaches the point of the second sighting. The buoy appears to have a bearing of 134 degrees at the first sighting, so the angle between the two legs of the triangle is (134 - 44 = 90) degrees. At the second sighting, the line between the boat and the ship is the hypotenuse of the triangle. Let it be y km long. The angle between the 15 km leg and the hypotenuse is 180 - 168 + 44 = 56 degrees. The third angle of the triangle is 90 - 56 = 34 degrees.



The tangent of the 56 degree angle is x/15.

x = 15 tan 56 deg = 22.238 km



The csc of the 34 deg angle is y/15.

y = 15 csc 34 deg or 15 / sin 34 deg = 26.824 deg

Answer 2

Your coordinate system has north (bearing 0°) straight up. Positive degrees are clockwise. The original position of the ship should be thought of as the origin.



The ship sails 15 km along a course 44° East of North. That is the first side of the triangle.



The second side is from the starting point to the buoy. It is 134° clockwise from north.

The angle between the first two sides is (134 - 44) = 90°.



The third side is from where the ship sailed to the buoy. The bearing from this position to the buoy is 168° clockwise of north. The angle between the first and second sides is the supplement of this plus the bearing of the first side (alternate interior angles)

(180 - 168) + 44 = 12 + 44 = 56°



The angle between the second and third sides is therefore

180 - 90 - 56 = 34°.



Now find the length of the second and third sides by the law of sines.



15/sin34° = x/sin56° = y/sin90°

x = 15(sin56°/sin34°) ≈ 22.238415 km

y = 15(sin90°/sin34°) = 15/sin34° ≈ 26.824375 km

Answer 3

What you need to do is start with a coordinate system inwhich the origin is the location of the ship. Draw a line 44 degrees up and 15 long. This will be one of the legs of the triangle. Also from the origin you are going to draw a triangle 134 degrees (which should go into the second coordinate. You don't know how long this line is so just call the line x. This will also be a leg of the triangle. At this point you chould be able to figure out the ange between from the 15 line to the x line. The angle is 134-44=90 degrees.



Now, we need to figure out one more angle and then we can use trigonometric functions (sine, cosine, and tangent) to figure out the lengths. What we are going to do is draw another set of imaginary axis at the other end of the 15 line. This is basically our new point of view (after sailing 15km). Now we should be able to draw another line at a 168 degree angle in the new coordinate system. Now we have the complete triangle and we can figure out the angle at the point in this new coordinate system. It is made of two angles we know. The first is from the third line (y) to the y-axis of the second coordinate system. That angle is 180-168=12 degrees. To figure out the amount of the rest of the angle, we have to use the rules of the angles with respect to paralell lines (I can't remember what they are called). The parallel lines are the y-axes of the two coordinate systems that we have and the relavant angles are created by the 15 line that joins the two. The 44 degrees on the bottom axis is the same as the angle from the top axis to the 15 line, so the rest of the angle in question is 44 degrees, so the total angle is 56 degrees.



Now we can figure out the lengths. Line x is opposite to the 56 degree angle and the 15 line is adjacent to the 56 degree angle. Therefore the tan(56)=x/15, so x=15tan(56)=22.2km. Line y is the hypotenous, so cos(56)=15/y, so y=15/cos(56)=26.8km.

Answer 4

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