Dr. J's Maths.com
Where the techniques of Maths
are explained in simple terms.

Trigonometry - Bearings (compass and true).
Test Yourself 1 - solutions.


 

Conversions from compass to true bearings. 1. Convert the following compass bearings to true bearings:
(i) South is 180°.

(ii) West is 270°.

(iii) North west is 315°.

(iv) North east is 045°.

 (v) North north east (NNE) is 022.5°

(vi) West south west is 247.5°

(vii) East 25° south is is 115°.

(viii) West 15° north is 285°.
Conversions from true to compass bearings. 2. Convert the following true bearings to compass bearings:

(i) 0° T is North.

(ii) 270° T is West.

(iii) 135° T is South-east.

(iv) 020° T is North 20° E.

(v) 210° T is West 30° South.

(vi) 310° T is South 40° East.

(vii) 157.5° is West-North-West.

(vii) 202.5° is North-South-West.
Bearings "looking back" FROM 3. The compass bearing of Point A from Point B is N38°E.

(i)

(ii) Point B from Point A is at a bearing of 218° T
or South 38° West
or East 52° South?
 

4. When the ship changes bearing to sail along a course at 240°, it has turned 60° from the N-S direction. Using complementary angles, the other two angles shown in red can be determined.

The ship is closest to point A when the line from point A is at right angles to its new direction. Hence the bearing of the ship from point A will be 90° + 60° = 150°
 

5. (i)

ALWAYS DRAW THE AXES AT THE OTHER POINT AND MARK IN THE COMPLEMENTARY ANGLE.

(ii) North 42° West of West 48° North.
  6.
Bearings in right angle triangles.

7. A ship travels 150 km in a north west direction.

(i)

(ii)

 

 
   
   
   
Bearings with other rules. 11. (i)

(ii)

Check the diagram:

We know one angle
(SPA = 50°) and there are 3 sides involved - so a cosine rule.
 

Three ships start from Port O.

  • Ship A sails at 60 nautical miles per hour at a bearing of 342°;
  • Ship B sails at 40 nautical miles per hour at a bearing of 128°;
  • Ship C sails at 30 nautical miles per hour at a bearing of 058°.

(i) Draw a diagram to indicate the position of each ship after 45 minutes have elapsed.

(ii) Calculate the diatance of Ship B from Ship A;

(iii) Find the bearing of ship C from Ship B.

Answer.(ii) AB = 72 nm.
(ii) Bearing of C from B is 338°T.
 

The diagram below shows three towns, A, W and B. A plane travelling from town A to town B must make a stop-over at town W. Town W is 120 km on a true bearing of 312° from town A. Town B, which is directly north of town A, is on a true bearing of 042° from town W.

(i) Draw a diagram to represent this information about the relative location of the three towns.

(ii) How much farther is the plane’s trip with the stop-over at W than if it were to fly directly from A to C? Express your answer to the nearest kilometre.


Answer.(ii) 253 - 179 = 74 km.
 

17. A course for a Championship Drone event requires competitors to fly from point A on a bearing of 050° for 1.2 km over cliffs with significant updrafts to point B. From point B they must then fly on a bearing of 230° over land for 2.8 km to point C.

(i) Draw a diagram to represent this information.

(ii) Find the distance from point C to point A, to the nearest 100 m.

(iii) What is the bearing of C from A?
 

18.

(i) ESE = 22.5° below direct east.
So the true bearing is
90° + 22.5° = 112.5°

(ii) Looking at me from the beach requires looking 22.5° from North. Hence the bearing to me FROM the beach is 22.5°.

 

(iii) From my position, I am looking 22.5° further than due South along the beach.

Hence the beach FROM my position is:
180° + 22.5° = 202.5°.

This is the same as S22.5°W.