Compassbearingexample

a. Draw a compass from the point and always face the north line.b. Moving clockwise, east will be 090°, south is 180°, and west is 270°. It will be between the easterly and southerly direction.c. Measuring the angle 135-degrees, roughly, the blue line will be 135°.

Example 3:A ship from the port travels for 60 km on the bearing of 120°. It continuous to travel and changes direction for another 50km on a bearing of 080°. How far is the ship from the starting point?

Bearing exampleswith answers

What are Bearings in Maths?A bearing measures the movement of an angle in a clockwise direction and always on the north line. The bearing of a point is the line joining the centre of the compass through the point measured in degrees in a clockwise way from the north direction. In navigation, bearings are used to express something about direction. It is also used to express in angle a particular landmark.

Bearings Maths questions and answers

Measuring BearingsExample:Imagine you are on an island in the middle of the ocean. You have a radio to ask for rescue. You received a message to turn 135 degrees. Where is 135 degrees?

Bearingquestions and answers pdf

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The three characteristics of bearings:a.The basis of a bearing is at the north directionb.It always measures clockwisec.It is written in 3-digit angles.

Example 2:The first ship leaves the port and travels 20 km on a bearing 0f 260°. At the same time, a second ship also leaves the same port and travels 14 km on a bearing of 190°. How far apart are the two ships?

Bearing examplesmaths

\( b^2 = a^2 + c^2 – 2ac  \cos B\)                           Substitute the given\( b^2 = {(50)}^2 + {(60)}^2 – 2(50)(60) \cos 140°\)\( b^2 = {(50)}^2 + {(60)}^2 – 2(50)(60) \cos 140°\)\( b = 103.42 km \)

We’ve formed a triangle. Now analyse the illustration:a. If the measure of angle A is 120°, then the measure of the remaining angle through the south is 60°.b. Let’s assume that from point A through the north the angle is 60°.c. Let’s say the current location of the ship is point C. ∠ABC = 60° + 80° = 140°.d.Remember, cosine law. It is used to find the measure of an angle or sides of a non right triangle. Let’s use \( b^2 = a^2 + c^2 – 2ac \cos B\).

a. Let’s label as point C the starting point or the port.b. The angle ACB = 250° + 190° = 60°c. From the port (C) to the location of the first ship (A) is 20 km. The second ship (B) from the port (C) is 14 km.d. We will now use the cosine law:\( c^2 = a^2 + b^2 – 2ab \cos C\)\( c^2 = 14^2 + 20^2 – 2(14)(20) \cos 60°\)\( c= 17.78 km\)