Bearing...2

Section 4.8

Can you set parts (a & b) for me? The N 32° W and N 68° W is not clear to me.

 

first draw a picture



(a) the bearing from A to C
angle A = 90°
drawing vertical line (represents N) at point A ad point C we see that angle A = 90° shows bearing from A to B is N 32° W , so 90°-32° =58°
so, bearing from A to C is N 58° E

(b) the distance from A to B.

we need inner angle of right triangle at C ad it is 54° ( see drawing)

tan 54° = AB/50
AB = 50tan(54) = 68.82 m

 

The "bearing" from one point to another is the angle from due north. Since the bearing from A to B is N 32 W the bearing from A to C is 90- 32= N 58 E.
Since the bearing from C to B is N 68 W, a line due north will cut the line from A to C at 32 degrees and the line from B to C at 68 degrees. A triangle from those two points to A will have two angles of 32 and 68 degrees so the angle at A is 180- 32- 68= 180- 100= 80 degrees. So AB is a leg on a triangle with angle 80 degrees and opposite leg has length 50 m.

 

Thank you. Like I said, I will watch a few video lessons to get a better grasp of this interesting but challenging topic. I assume this is something a captain of a ship must be familiar with. Yes?

 

I will practice a few more samples after watching several video lessons.

 

Nice picture. How did you draw this triangle?

 

Normally, the right angle is angle C. Here, the right angle or 90 degrees is listed as angle A. Am I right?

 

right

 

Cool. I am still a bit rusty in terms of bearing problems. I will post 3 or 4 more today and then move in to Simple Harmonic Motion before saying goodbye to Chapter 4.