From a location A , The bearing of two places B and C are N58 degree W and N32 degree E respectively. If the bearing of C from B is N80 degree E and AB = 120m, find AC and the bearing of B from C?

Clarks

1 year ago

Given:

From point A:

Bearing of B is N 58° W → This means: 58° west of due north.

Bearing of C is N 32° E → This means: 32° east of due north.


Bearing of C from B is N 80° E → 80° east of due north.

Distance AB = 120 m


We are to find:

1. AC


2. Bearing of B from C




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Step 1: Draw the diagram

Draw point A. From A:

Draw a line going northwest (N58°W) to point B.

Draw a line going northeast (N32°E) to point C.


The angle between these two lines at point A = 58° + 32° = 90°
(Why? Because the bearings are measured from the north line; they’re on opposite sides — west and east.)

So angle BAC = 90°

Now, triangle ABC is formed with:

AB = 120 m

Angle BAC = 90°

Need to find AC


Step 2: Use the Law of Sines or Cosines

We don’t need to use cosine or sine law yet because we have a right-angled triangle at A (since angle BAC = 90°).

Now we draw angle at B.

We're told the bearing of C from B is N 80° E. That means:

From B, C is located 80° east of due north.

But remember, from A, B is at N58°W, and C is at N32°E.

So angle at B, i.e., angle ABC, is 180° - 80° = 100° (This is because bearing of C from B is measured clockwise from north, so the internal triangle angle at B = 180° – bearing)


Now, in triangle ABC:

Angle BAC = 90°

Angle ABC = 100°

So angle ACB = 180° - 90° - 100° = -10°


Wait, that’s not possible. So something’s wrong. Let’s fix that.
Recheck the angle logic carefully

From point A:

B is at N 58° W

C is at N 32° E


So the total angle at A = 58° + 32° = 90° (we keep this)

Now, the bearing of C from B is N 80° E. That’s from B to C.

Now, from point B:

North line is drawn

80° toward east gives us the direction from B to C


Now we can use triangle ABC with:

AB = 120 m

angle at A = 90°

angle at B = 80° (since bearing from B to C is N80°E, which makes angle ABC = 80°)


So we have:

angle A = 90°

angle B = 80°

angle C = 10° (since sum = 180°)

l

Step 3: Use Law of Sines to find AC

In triangle ABC:

Angle A = 90°

Angle B = 80°

Angle C = 10°

AB = 120 m (side between angle B and angle C)


Law of sines:

\frac{AB}{\sin(C)} = \frac{AC}{\sin(B)}

Plug in values:

\frac{120}{\sin(10°)} = \frac{AC}{\sin(80°)}

Now calculate:

sin(10°) ≈ 0.1736

sin(80°) ≈ 0.9848


\frac{120}{0.1736} = \frac{AC}{0.9848}

691.2 = \frac{AC}{0.9848}

Now solve for AC:

AC = 691.2 \times 0.9848 ≈ 680.7 \, \text{meters}


Step 4: Find bearing of B from C

We already know:

From B, the bearing of C is N 80° E

So from C, the bearing of B is exactly opposite


To reverse a bearing, add or subtract 180°:

80° + 180° = 260°


Now express 260° in compass notation:

260° is between 180° and 270°, so it's in the southwest quadrant.

How far is 260° from the south line (which is 180°)?
260° - 180° = 80°


So the bearing is: S 80° W

Final Answers:

1. AC ≈ 680.7 meters
2. Bearing of B from C = S 80° W