Bearings Worked ExamplesIGCSE Maths 0580

A back bearing differs from the forward bearing by $180^{\circ}$. If the original is less than $180^{\circ}$, add.

(b) The original is more than $180^{\circ}$, so subtract — otherwise the result would exceed $360^{\circ}$.

Bearings are measured clockwise from north. A bearing of $130^{\circ}$ points south-east, so the ship moves east and south.

The angle from the north line is $130^{\circ}$; the angle from the south line is $180^{\circ} - 130^{\circ} = 50^{\circ}$.

East component uses sine of the bearing angle.

South component uses cosine of the angle from south, $50^{\circ}$.

(a) The bearing turns from $040^{\circ}$ to $130^{\circ}$, a change of direction of $90^{\circ}$.

The two legs are therefore perpendicular, so triangle $PQR$ is right-angled at $Q$.

(b) Use Pythagoras rather than the cosine rule — it is exact and faster.

Square root.

Work with total components from $P$. East: $12\sin 40^{\circ} + 18\sin 130^{\circ}$.

North: $12\cos 40^{\circ} + 18\cos 130^{\circ}$.

So $R$ is east and slightly south of $P$. The bearing of $R$ from $P$ lies between $090^{\circ}$ and $180^{\circ}$.

Bearing of $R$ from $P$ is $90^{\circ}$ plus that angle.

The return bearing is the back bearing — add $180^{\circ}$.

(a) North-east is exactly halfway between north ($000^{\circ}$) and east ($090^{\circ}$).

(b) The walker goes due south from $Y$ to $Z$, a bearing of $180^{\circ}$. The bearing of $Y$ from $Z$ is the reverse.

Draw the north line at $H$ and mark the $055^{\circ}$ angle clockwise from it.

The northward distance is adjacent to the bearing angle, with $15\text{ km}$ as the hypotenuse, so use cosine.

Evaluate.

(a) The bearing of $C$ from $A$ is the back bearing of $292^{\circ}$. It is over $180^{\circ}$, so subtract.

(b) Both bearings from $A$ are measured clockwise from the same north line, so the angle between $AB$ and $AC$ is their difference.

(c) $AB = AC$, so triangle $ABC$ is isosceles and its base angles are equal.

At $B$, the bearing of $A$ is the back bearing $048^{\circ} + 180^{\circ} = 228^{\circ}$. The diagram shows that $BC$ points closer to south than $BA$ does, so turn anticlockwise from $BA$ by $58^{\circ}$.

(d) This is the back bearing of the answer to (c).

(a) The angle between $PL$ and the ship's course is the difference between the two bearings.

The shortest distance from $L$ to the course is along the perpendicular. Let it meet the course at $X$. Triangle $PXL$ has a right angle at $X$, with hypotenuse $PL = 12$, and $LX$ is opposite the $38^{\circ}$ angle.

(b) The ship is closest at $X$. The distance sailed, $PX$, is adjacent to the $38^{\circ}$ angle.

Time is distance divided by speed. Convert hours to minutes.

(c) At $X$, the direction to $L$ is at $90^{\circ}$ to the course. The diagram shows $L$ to the left of the course (on the north side), so turn $90^{\circ}$ anticlockwise from $070^{\circ}$.

(a) At $A$, both bearings are measured from the same north line.

At $B$, the bearing of $A$ is the back bearing $100^{\circ} + 180^{\circ} = 280^{\circ}$. The angle between $BA$ and $BS$ is the difference from $335^{\circ}$.

Angles in a triangle sum to $180^{\circ}$. Two angles are equal, so the triangle is isosceles, with $AS = AB = 10\text{ km}$ (the sides opposite the equal angles).

(b) $BS$ is opposite the $70^{\circ}$ angle, and $AB = 10$ is opposite a $55^{\circ}$ angle. Use the sine rule.

(c) Work in triangle $ASX$. At $S$, the bearing of $A$ is $030^{\circ} + 180^{\circ} = 210^{\circ}$ and due south is $180^{\circ}$, so the angle between them is $30^{\circ}$. At $A$, $X$ lies on $AB$, so angle $SAX$ is still $70^{\circ}$.

$SX$ is opposite the $70^{\circ}$ angle, and $AS = 10$ (from part (a)) is opposite the $80^{\circ}$ angle.

Sketch $P$ and $Q$ with a north line at each. The north lines are parallel, so the two bearings differ by exactly $180^{\circ}$. Try the case where the bearing of $Q$ from $P$ is the smaller one, so the back bearing is $180^{\circ}$ more.

Collect the $x$ terms on one side and the numbers on the other.

Check both bearings. They must lie between $000^{\circ}$ and $360^{\circ}$ and differ by $180^{\circ}$.

The other case, $5x - 50 = (2x + 10) - 180$, gives $3x = -120$ and $x = -40$. Then the bearing of $Q$ from $P$ would be $-70^{\circ}$, which is impossible, so $x = 80$ is the only answer.

(a) Put the centre of the protractor on $H$ with the zero line along the north line. Measure clockwise from north to the line $HJ$. $J$ is east and a little south of $H$, so the bearing must be between $090^{\circ}$ and $180^{\circ}$.

Check that you used the right scale on the protractor. Reading the other scale gives $180^{\circ} - 112^{\circ} = 68^{\circ}$, which points north of east and is clearly wrong.

(b) Set the compasses to $9\text{ km}$ using the scale bar and draw an arc with centre $H$. Then set them to $7\text{ km}$ and draw an arc with centre $J$. The arcs cross at two points, one on each side of $HJ$. $M$ is the crossing point south of $HJ$. Leave both arcs showing.

(c) Join $H$ to $M$ and measure clockwise from the north line at $H$. You can check the reading by calculation. $HJ$ measures $12\text{ km}$ on the scale, so the cosine rule in triangle $HJM$ gives angle $JHM$.

$M$ is south of $HJ$, so $HM$ is clockwise from $HJ$.

(a) Draw a north line at $T$. The bearing of $S$ from $T$ is the back bearing of $126^{\circ}$.

From the diagram, $TU$ is $41^{\circ}$ clockwise from $TS$ (closer to north).

(b) Two sides and the angle between them are known, so use the cosine rule. Work out $2 \times 7.2 \times 11.8 \cos 41^{\circ}$ as a single term before subtracting.

Take the square root and write down more figures than the value you are asked to show.

(c) $TU = 11.8$ is opposite angle $TSU$, and $SU$ is opposite the $41^{\circ}$ angle. Use the sine rule with the unrounded $SU$.

The calculator gives the acute angle, but angle $TSU$ is obtuse. Use the supplement.

At $S$, the bearing of $T$ is $126^{\circ}$. $SU$ is anticlockwise from $ST$ (closer to north), so subtract.

(d) The third leg goes from $U$ to $S$, which is the back bearing of the answer to (c).