GCSE Maths Loci, Bearings and Vectors (Higher): check-up and exam-style questions

Key words. A scale compares a length on a map or drawing with the real length. A scale of 1 : 50 000 means that 1 cm on the map is 50 000 cm in real life. A ratio compares two amounts in the same units.

Method. Multiply the length on the map by the scale number to get centimetres, then change to sensible units.

1. Say what the scale means. A scale of 1 : 50 000 means that every 1 cm on the map stands for 50 000 cm in real life.
   1 cm on the map = 50 000 cm real
2. Multiply the map length by 50 000. The real distance is 50 000 times as long as the length on the map.
   6 × 50 000 = 300 000 cm
3. Change the centimetres into km. There are 100 cm in 1 m and 1000 m in 1 km, so there are 100 000 cm in 1 km. Divide by 100 000.
   300 000 cm ÷ 100 000 = 3 km

Check. Go backwards: 3 km = 300 000 cm and 300 000 ÷ 50 000 = 6 cm, the length we started with. ✓

Watch out. Multiply by the scale number to go from the map to real life. Divide to go the other way. Change the units at the end.

Key words. A scale compares a length on a map or drawing with the real length. A scale of 1 : 50 000 means that 1 cm on the map is 50 000 cm in real life. A ratio compares two amounts in the same units.

Method. Write the map length and the real length in the same unit (cm), make a ratio, and divide so the first number is 1.

1. Change the real length into cm. Both parts of a ratio must be in the same unit. 1 km = 1000 m = 100 000 cm, so multiply by 100 000.
   2 km × 100 000 = 200 000 cm
2. Write the ratio map : real. The scale compares the map length with the real length, in that order.
   1 : 200 000
3. The first number is already 1. So the scale is complete.
   Scale = 1 : 200 000

Check. 1 cm on the map stands for 200 000 cm = 2 km in real life. So 1 cm on the map should be 2 km. ✓

Watch out. Change to the same units first. A ratio like 2 : 4 km is wrong because the units do not match.

Key words. A bearing is an angle that gives a direction. It is measured from north, clockwise, and written with three figures (for example 045° or 270°). A back bearing is the bearing of the way back along the same line.

Method. The bearing of A from B is the back bearing of the bearing of B from A: add or subtract 180° so the answer stays between 0° and 360°.

1. Draw a sketch with a north line at each point. The two north lines both point the same way, so they are parallel. The line AB crosses both of them.
   Bearing of B from A = 065°
2. Find the angle at B between the north line and BA. Angles between parallel lines and on the same side of the crossing line are co-interior and add up to 180°.
   180° − 065° = 115°
3. Turn it into a bearing. A bearing is measured clockwise from north, but this angle is measured the other way round. Take it away from a full turn of 360°.
   360° − 115° = 245°
4. Write it with three figures and check with the shortcut. Going back is the exact opposite way, which is half a turn (180°) different from the forward bearing.
   65° + 180° = 245°. Back bearing = 245°

Check. Reverse it again: 245° − 180° = 65°. ✓

Watch out. Always draw the north line at the point you are measuring from. If the forward bearing is less than 180° add 180°; if it is more than 180° subtract 180°.

Key words. A locus (plural loci) is the set of all the points that obey a rule. Equidistant means the same distance away.

Method. “The same distance from two points” is one of the four basic loci.

1. Recall the rule. Points equidistant from A and B lie on the line that cuts AB in half at right angles.
   Perpendicular bisector of AB
2. Write the answer. You construct it with equal arcs from A and from B.
   The perpendicular bisector of AB.

Check. The midpoint of AB is on this line. ✓

Key words. A locus (plural loci) is the set of all the points that obey a rule. Equidistant means the same distance away.

Method. A fixed distance from a fixed point gives a circle.

1. Recall the rule. All points at the same distance from P form a circle.
   Centre P
2. State the radius. The fixed distance is the radius.
   A circle with centre P and radius 4 cm.

Check. Any point on the circle is 4 cm from P. ✓

Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.

Method. To add column vectors, add the top numbers and add the bottom numbers separately.

1. Write the two column vectors. Each column vector has a top number (across) and a bottom number (up).
   a + b = 2
   −3 + −4
   5
2. Add the top numbers. To add vectors, add the matching numbers: top with top.
   2 + −4 = −2
3. Add the bottom numbers. Now do the same with the bottom with bottom.
   −3 + 5 = 2
4. Write the answer as a column vector. Put the new top number over the new bottom number.
   a + b = −2
   2

Check. On a grid: start at the origin, go 2 right and 3 down, then follow b. You arrive 2 left and 2 up of the start. ✓

Watch out. Take care with minus signs: subtracting a negative number adds. For example 3 − (−2) = 5.

Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.

Method. Multiply the top number and the bottom number of the vector by the scalar.

1. Write the multiplication. A scalar multiple means the vector is stretched (or shrunk) by the number in front.
   3a = 3 × 2
   −3
2. Multiply the top number. Multiply every number in the column vector by the scalar.
   3 × 2 = 6
3. Multiply the bottom number. Do the same with the bottom number.
   3 × (−3) = −9
4. Write the answer. Put the two results in a column.
   3a = 6
   −9

Check. The new vector is 3 times as long as a and points the same way. ✓

Watch out. Multiply both numbers, not just the top one.

Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.

Method. Subtract the coordinates of the start point from the coordinates of the end point.

1. Write the two points. A column vector from A to B says how far to move across and up to get from A to B.
   A = (1, 2) and B = (5, 5)
2. Find the across move. Subtract the x-coordinates: end minus start.
   5 − 1 = 4
3. Find the up move. Subtract the y-coordinates: end minus start.
   5 − 2 = 3
4. Write the column vector. Across on top, up underneath.
   →AB = 4
   3

Check. Start at A and add the vector: (1 + 4, 2 + 3) = (5, 5) = B. ✓

Watch out. It is end minus start. If you swap them you get the vector going the other way, which has the opposite signs.

Key words. A bearing is an angle that gives a direction. It is measured from north, clockwise, and written with three figures (for example 045° or 270°). A back bearing is the bearing of the way back along the same line.

Method. The bearing of A from B is the back bearing of the bearing of B from A: add or subtract 180° so the answer stays between 0° and 360°.

1. Draw a sketch with a north line at each point. The two north lines are parallel. The line AB crosses both of them.
   Bearing of B from A = 300°
2. Find the angle between the north-south line at A and AB. The bearing is more than 180°, so the line has gone past south. The extra angle past south is the bearing minus 180°.
   300° − 180° = 120°
3. Use alternate angles at B. Alternate angles (a Z shape) between parallel lines are equal. So the angle at B between the north line and BA is the same size, and it is measured clockwise from north.
   Angle = 120°
4. Write it with three figures and check with the shortcut. Going back is half a turn different. Because the forward bearing is more than 180°, subtract 180°.
   300° − 180° = 120°. Back bearing = 120°

Check. Reverse it again: 120° + 180° = 300°. ✓

Watch out. Always draw the north line at the point you are measuring from. If the forward bearing is less than 180° add 180°; if it is more than 180° subtract 180°.

Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.

Method. Find a route from O to M using vectors you know. Moving against an arrow gives the negative of that vector.

1. Choose a route. Go from O to M along paths whose vectors we know or can work out. Add the vectors on the way.
   →OM = →OA + →AM
2. Work out OA. OA is given as a.
   →OA = a
3. Work out AM. M is the midpoint of AB, so AM = ½AB. AB = b − a.
   →AM = 12b − 12a
4. Add the parts and simplify. Collect the a terms together and the b terms together.
   →OM = (a) + (12b − 12a)
   = 12a + 12b

Check. Try a different route from O to M, or use numbers (say a = (2, 0) and b = (0, 3)): both routes give the same vector. ✓

Watch out. Going the opposite way along a vector reverses its sign: AO = −OA. Always say which way each arrow points.

Key words. A scale compares a length on a map or drawing with the real length. A scale of 1 : 50 000 means that 1 cm on the map is 50 000 cm in real life. A ratio compares two amounts in the same units.

Method. Multiply the length on the map by the scale number to get centimetres, then change to sensible units.

1. Say what the scale means. A scale of 1 : 25 000 means that every 1 cm on the map stands for 25 000 cm in real life.
   1 cm on the map = 25 000 cm real
2. Multiply the map length by 25 000. The real distance is 25 000 times as long as the length on the map.
   8.4000000000000003552713678800500929355621337890625 × 25 000 = 210000.0000000000088817841970012523233890533447265625 cm
3. Change the centimetres into km. There are 100 cm in 1 m and 1000 m in 1 km, so there are 100 000 cm in 1 km. Divide by 100 000.
   210000.0000000000088817841970012523233890533447265625 cm ÷ 100 000 = 2.100000000000000088817841970012523233890533447265625 km

Check. Go backwards: 2.100000000000000088817841970012523233890533447265625 km = 210000.0000000000088817841970012523233890533447265625 cm and 210000.0000000000088817841970012523233890533447265625 ÷ 25 000 = 8.4000000000000003552713678800500929355621337890625 cm, the length we started with. ✓

Watch out. Multiply by the scale number to go from the map to real life. Divide to go the other way. Change the units at the end.

Key words. To construct means to draw accurately using only a ruler and a pair of compasses (no protractor). The construction arcs are the curved lines the compasses leave; you must not rub them out.

Method. Use equal arcs from A and B, more than 3.5 cm, then join the places where they cross.

1. Draw AB. Use a ruler.
   AB = 7 cm
2. Draw arcs from A and B. Same radius, more than 3.5 cm. They cross above and below AB.
   Arcs from A, then from B
3. Join the crossing points. Leave the arcs.

Check. The line crosses AB at its midpoint, 3.5 cm from A, at 90°. ✓

Watch out. Marks: 1 for two correct pairs of arcs, 1 for the line.

Key words. A locus (plural loci) is the set of all the points that obey a rule. Equidistant means the same distance away.

Method. Draw each rule as a boundary line, then shade the region that obeys all the rules.

1. Look at the plan and list the rules. Each rule gives a locus. The answer must obey every rule at the same time.
   Rule 1: within 6 m of C
   Rule 2: nearer to C than to D
   Rule 3: at least 1 m from side BC
2. Draw rule 1: within 6 m of C. Set the compasses to 6 squares. Put the point on C and draw an arc inside the plan.
3. Draw rule 2: nearer to C than to D. Construct the perpendicular bisector of CD and shade the side containing C.
4. Draw rule 3: at least 1 m from side BC. Draw a line parallel to BC and 1 square to its left. Shade the side away from BC.
5. Shade the region that obeys all the rules. Points that are in the right place for every rule are where the regions overlap.

Check. Pick a point in the shaded region, for example (6.2, 2.6) measured from the bottom left corner of the plan. Test each rule:
distance from C = 3.7 m, which is at most 6 m ✓
distance from C = 3.7 m and from D = 6.6 m, so it is closer to C ✓
distance from BC = 2.8 m, which is at least 1 m ✓

Watch out. A point must pass every test. Shade only the overlap, not all of each region separately.

Key words. A bearing is an angle that gives a direction. It is measured from north, clockwise, and written with three figures (for example 045° or 270°). A back bearing is the bearing of the way back along the same line.

Method. Draw the journey and mark the right angle where the ship turns. Use Pythagoras for a distance and trigonometry for an angle, then turn the angle into a bearing.

1. Draw a sketch and put a north line at A and at B. A bearing is always measured from north, so we need a north line at every point we measure from.
   A to B: 12 km on a bearing of 040°
   B to C: 5 km on a bearing of 130°
2. Find angle ABC. The ship changes direction from 040° to 130°. The bearings differ by 90°, so the ship turns right through a quarter turn and the path makes a right angle at B.
   130° − 040° = 90°
   Angle ABC = 90°. Triangle ABC is right-angled.
3. Use Pythagoras to find AC. AC is opposite the right angle, so it is the hypotenuse: add the squares, then square root.
   AC² = 12² + 5² = 144 + 25 = 169
   AC = √169 = 13.0 km
4. Find angle BAC with trigonometry. In the right-angled triangle we know the side opposite the angle (BC = 5) and the side next to it (AB = 12). Opposite and adjacent means tan.
   tan(angle BAC) = 5 ÷ 12
   angle BAC = tan⁻¹(5 ÷ 12) = 22.6°
5. Add to the bearing of B from A. C is further round clockwise from north than B is, so add the angle to 040°.
   40° + 22.6° = 62.6°
   Bearing of C from A = 063° (to the nearest degree)

Check. Using coordinates from A (east, north): B = (7.71, 9.19) and C = (11.54, 5.98), so C is 13.0 km from A on a bearing of 063°. ✓

Watch out. The angle found with trigonometry is inside the triangle. You must add it to (or subtract it from) a bearing before you answer.

Key words. A bearing is an angle that gives a direction. It is measured from north, clockwise, and written with three figures (for example 045° or 270°). A back bearing is the bearing of the way back along the same line.

Method. The bearing of A from B is the back bearing of the bearing of B from A: add or subtract 180° so the answer stays between 0° and 360°.

1. Draw a sketch with a north line at each point. The two north lines both point the same way, so they are parallel. The line AB crosses both of them.
   Bearing of B from A = 125°
2. Find the angle at B between the north line and BA. Angles between parallel lines and on the same side of the crossing line are co-interior and add up to 180°.
   180° − 125° = 55°
3. Turn it into a bearing. A bearing is measured clockwise from north, but this angle is measured the other way round. Take it away from a full turn of 360°.
   360° − 55° = 305°
4. Write it with three figures and check with the shortcut. Going back is the exact opposite way, which is half a turn (180°) different from the forward bearing.
   125° + 180° = 305°. Back bearing = 305°

Check. Reverse it again: 305° − 180° = 125°. ✓

Watch out. Always draw the north line at the point you are measuring from. If the forward bearing is less than 180° add 180°; if it is more than 180° subtract 180°.

Key words. To construct means to draw accurately using only a ruler and a pair of compasses (no protractor). The construction arcs are the curved lines the compasses leave; you must not rub them out.

Method. Draw the longest side as a base, then find the third corner with two arcs. Measure the angle with a protractor.

1. Draw the base AB. Choose one side as the base and draw it with a ruler.
   AB = 6 cm
2. Draw an arc from A. C is 5 cm from A, so every possible position of C is on an arc of radius 5 cm centred on A.
   Compasses set to 5 cm, point on A.
3. Draw an arc from B. C is also 4 cm from B. Where the two arcs cross is the only point that is the right distance from both A and B.
   Compasses set to 4 cm, point on B. The arcs cross at C.
4. Join AC and BC. Use a ruler. Keep the arcs.
5. Measure angle C. The biggest angle is opposite the longest side (6 cm), which is angle C. Use a protractor.
   Angle C = 82.8° to 1 decimal place, so about 83°

Check. Cosine rule: cos C = (4² + 5² − 6²) ÷ (2 × 4 × 5) = 0.1250, so C = 82.8°. Your measurement should be within 1° of this. ✓

Watch out. The two arcs must have the radii of the two sides that meet at C. Do not swap them.

Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.

Method. Multiply each vector by its scalar first, then add or subtract the top numbers and the bottom numbers.

1. Write each vector as a column. We can only add and subtract once everything is in column form.
   2a − b = 2 × 3
   −1 + −1 × −2
   4
2. Do the scalar multiplications first. Multiply every number in each vector by its scalar (like BIDMAS: multiply before adding).
   6
   −2 + 2
   −4
3. Add the top numbers. Add the matching numbers: top with top.
   6 + 2 = 8
4. Add the bottom numbers. Then bottom with bottom.
   (−2) + (−4) = −6
5. Write the answer.
   2a − b = 8
   −6

Check. Check the top numbers again: 6 + 2 = 8. ✓

Watch out. Do not forget to multiply both numbers, and watch the signs when you subtract a vector.

Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.

Method. Add the vector to each vertex, then find the length of the translation with Pythagoras.

1. Translate A. Add the vector.
   (1 + (−3), 1 + 2) = (−2, 3)
2. Translate B and C. The same vector for each.
   (4 + (−3), 1 + 2) = (1, 3) and (4 + (−3), 3 + 2) = (1, 5)
3. Find the distance. The distance each point moves is the magnitude of the vector.
   |vector| = √((−3)² + 2²) = √13
4. Write the answers.
   A′(−2, 3), B′(1, 3), C′(1, 5); distance √13

Check. √13 = 3.6… units. ✓

Watch out. A translation moves every point by the same distance.

Key words. A scale compares a length on a map or drawing with the real length. A scale of 1 : 50 000 means that 1 cm on the map is 50 000 cm in real life. A ratio compares two amounts in the same units.

Method. Change each plan length to a real length in metres, then multiply to find the area.

1. Find what 1 cm stands for. 1 : 200 means 1 cm is 200 cm = 2 m.
   1 cm on the plan = 2 m
2. Real length. Multiply by 2.
   5 × 2 = 10 m
3. Real width.
   3 × 2 = 6 m
4. Area. Area = length × width.
   10 × 6 = 60 m²

Check. On the plan the area is 15 cm², and 1 cm² stands for 2 × 2 = 4 m²: 15 × 4 = 60. ✓

Watch out. Do not multiply the area by 200.

Key words. A bearing is an angle that gives a direction. It is measured from north, clockwise, and written with three figures (for example 045° or 270°). A back bearing is the bearing of the way back along the same line.

Method. Find the angle inside the triangle at the turning point from the two bearings. Then use the cosine rule for the missing side and the sine rule for an angle. Finish by turning the angle into a bearing.

1. Draw a sketch with a north line at A and B. We measure every bearing from a north line.
   A to B: 10 km, bearing 035°
   B to C: 7 km, bearing 110°
2. Find the bearing of A from B. This back bearing gives the direction of the line B to A. Add or subtract 180°.
   35° + 180° = 215°
3. Find angle ABC. It is the difference between the bearings of A and C as seen from B (take the smaller angle round).
   215° − 110° = 105°
   Angle ABC = 105°
4. Use the cosine rule to find AC. We know two sides (10 and 7) and the angle between them (105°), so use a² = b² + c² − 2bc cos A.
   AC² = 10² + 7² − 2 × 10 × 7 × cos 105°
   AC² = 185.23
   AC = 13.6 km
5. Use the sine rule to find angle BAC. We now know a side and the angle opposite it (AC and angle B), and we want the angle opposite BC (7 km).
   sin A ÷ 7 = sin 105° ÷ 13.610
   sin A = 7 × sin 105° ÷ 13.610 = 0.4968
   A = sin⁻¹(0.4968) = 29.8°
6. Add to (or subtract from) the bearing AB. C is clockwise from B as seen from A, because the ship turned right at B.
   35° + 29.8° = 64.8°
   Bearing of C from A = 065° to the nearest degree

Check. Using coordinates from A: B = (5.74, 8.19) and C = (12.31, 5.80), so C is 13.6 km from A on a bearing of 065°. ✓

Watch out. Use the angle inside the triangle in the cosine rule, not the turn angle. They add up to 180°.

Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.

Method. Find AB by the route A to O to B. Then OP = OA + (2/3) AB.

1. Part (a). Go from A back to O, then to B.
   →AB = −a + b = b − a
2. Find AP. AP : PB = 2 : 1, so AP is 2 of the 3 equal parts of AB.
   →AP = 23(b − a)
3. Find OP. Go from O to A, then along AP.
   →OP = a + 23(b − a) = 13a + 23b

Check. With a = (3, 0) and b = (0, 3): A = (3, 0), B = (0, 3), P = (1, 2). P is 2/3 of the way from A to B. ✓

Watch out. Marks: 1 for AB, 1 for the fraction of AB, 1 for the method, 1 for the answer.

Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.

Method. Find a route from P to Q using vectors you know. Moving against an arrow gives the negative of that vector.

1. Choose a route. Go from P to Q along paths whose vectors we know or can work out. Add the vectors on the way.
   →PQ = →PO + →OQ
2. Work out PO. P is on OA with OP = ⅔a, so PO = −⅔a.
   →PO = −23a
3. Work out OQ. Q is on OB with OQ = ⅔b.
   →OQ = 23b
4. Add the parts and simplify. Collect the a terms together and the b terms together.
   →PQ = (−23a) + (23b)
   = 23b − 23a
5. Compare with AB. AB = b − a.
   →PQ = 23b − 23a = ⅔(b − a) = ⅔→AB
6. Write the conclusion. PQ is a scalar multiple of AB.
   PQ is parallel to AB.

Check. Try a different route from P to Q, or use numbers (say a = (2, 0) and b = (0, 3)): both routes give the same vector. ✓

Watch out. Going the opposite way along a vector reverses its sign: AO = −OA. Always say which way each arrow points.

Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.

Method. Write OX in two different ways (once along each line), then equate the coefficients of a and b to find the two unknown fractions.

1. Write AN and BM. Use the route A → O → N for AN, and B → O → M for BM.
   →AN = −a + 13b = 13b − a
   →BM = −b + 12a = 12a − b
2. Describe X using AN. X is on AN, so AX is some multiple λ of AN. Go from O to A and then along AN.
   →OX = →OA + λ→AN = a + λ(13b − a) = (1 − λ)a + 13λb
3. Describe X using BM. X is also on BM, so BX is some multiple μ of BM. Go from O to B and then along BM.
   →OX = →OB + μ→BM = b + μ(12a − b) = 12μa + (1 − μ)b
4. Compare the coefficients. a and b are not parallel, so the a parts must match and the b parts must match.
   1 − λ = 12μ   (1)
   13λ = 1 − μ   (2)
5. Solve the simultaneous equations. Rearrange (1) to get λ = 1 − 12μ, then substitute into (2) and collect the μ terms.
   13(1 − 12μ) = 1 − μ
   13 − 16μ = 1 − μ
   μ − 16μ = 1 − 13
   56μ = 23
   μ = 45
   λ = 1 − 12 × 45 = 35
6. Find OX. Substitute λ = 35 back into the expression from AN.
   →OX = (1 − 35)a + 13 × 35b = 25a + 15b

Check. Use the BM route: 12 × 45 = 25 for a and 1 − 45 = 15 for b. This agrees with 25a + 15b. ✓

Watch out. Both routes must give the same vector. Equating coefficients only works because a and b are not parallel.

Key words. A bearing is an angle that gives a direction. It is measured from north, clockwise, and written with three figures (for example 045° or 270°). A back bearing is the bearing of the way back along the same line.

Method. Draw the journey and mark the right angle where the ship turns. Use Pythagoras for a distance and trigonometry for an angle, then turn the angle into a bearing.

1. Draw a sketch and put a north line at A and at B. A bearing is always measured from north, so we need a north line at every point we measure from.
   A to B: 300 km on a bearing of 090°
   B to C: 400 km on a bearing of 180°
2. Find angle ABC. The ship changes direction from 090° to 180°. The bearings differ by 90°, so the ship turns right through a quarter turn and the path makes a right angle at B.
   180° − 090° = 90°
   Angle ABC = 90°. Triangle ABC is right-angled.
3. Use Pythagoras to find AC. AC is opposite the right angle, so it is the hypotenuse: add the squares, then square root.
   AC² = 300² + 400² = 90 000 + 160 000 = 250 000
   AC = √250 000 = 500 km
4. Find angle BAC with trigonometry. In the right-angled triangle we know the side opposite the angle (BC = 400) and the side next to it (AB = 300). Opposite and adjacent means tan.
   tan(angle BAC) = 400 ÷ 300
   angle BAC = tan⁻¹(400 ÷ 300) = 53.1°
5. Add to the bearing of B from A. C is further round clockwise from north than B is, so add the angle to 090°.
   90° + 53.1° = 143.1°
   Bearing of C from A = 143° (to the nearest degree)
6. Find the back bearing. The bearing of A from C is the opposite direction to the bearing of C from A, so add or subtract 180°.
   143° + 180° = 323°
   Bearing of A from C = 323°

Check. Using coordinates from A (east, north): B = (300.00, 0.00) and C = (300.00, −400.00), so C is 500 km from A on a bearing of 143°. ✓

Watch out. The angle found with trigonometry is inside the triangle. You must add it to (or subtract it from) a bearing before you answer.

Key words. To construct means to draw accurately using only a ruler and a pair of compasses (no protractor). The construction arcs are the curved lines the compasses leave; you must not rub them out.

Method. Compare the perpendicular with a slanted line using a right-angled triangle.

1. Draw the picture. Let F be the foot of the perpendicular from P, and let X be any other point on the line.
   PF is perpendicular to the line, so angle PFX = 90°
2. Use the hypotenuse. In triangle PFX, PX is opposite the right angle, so PX is the hypotenuse.
   PX is the longest side
3. Conclude. The hypotenuse is longer than the shorter side PF.
   PX > PF, so the perpendicular PF is the shortest distance.

Check. If PF = 3 and FX = 4, then PX = 5, which is longer than 3. ✓

Watch out. Marks: 1 for the right-angled triangle, 1 for hypotenuse longest.