Bearing...2

[nycmathguy]

Section 4.8

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

 

[MathLover1]

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

 

[HallsofIvy]

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.

 

[nycmathguy]

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

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?

 

[nycmathguy]

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.

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

 

[nycmathguy]

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.

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

Nice picture. How did you draw this triangle?

 

[nycmathguy]

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

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

 

[MathLover1]

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

right

 

[nycmathguy]

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.