Bearings and geometric modelling

(a) 23 km

1. Each centimetre is 5 km: $4.6 \times 5 = 23$.

• M1 Writing $4.6 \times 5$.
• A1 Correct answer: 23 km.

(b) $135^\circ$

1. South-east is half way between east ($090^\circ$) and south ($180^\circ$), measured clockwise from north.

• B1 Correct answer: $135^\circ$.

(a) $254$°

1. $$74+180$$
2. Reverse bearings differ by $180^\circ.$
3. $074^\circ$ + $180^\circ$ = $254^\circ.$

• P1 Establishing $74+180$ or an equivalent valid method.
• A1 Correct answer: $254$°

(a) $225$°

1. A reverse bearing differs by 180 degrees.
   $$45+180$$
2. Therefore $225$°.

• M1 A reverse bearing differs by 180 degrees.
• A1 Correct answer: $225$°

(b) A fixed north reference makes each direction unambiguous; bearings increase clockwise from north.

1. A fixed north reference makes each direction unambiguous; bearings increase clockwise from north.

• C1 Correct conclusion with supporting reasoning: A fixed north reference makes each direction unambiguous; bearings increase clockwise from north.

(a) $15$ km

1. East and north displacements are perpendicular.
   $$\sqrt{9^{2}+12^{2}}$$
2. Therefore $15$ km.

• P1 East and north displacements are perpendicular.
• A1 Correct answer: $15$ km

(b) $217$°

1. The return direction is west of south; find its angle from south.
2. Add the acute angle to a south bearing.
   $$180+36.8698976458$$
3. Therefore $217$°.

• P1 The return direction is west of south; find its angle from south.
• P1 Add the acute angle to a south bearing.
• A1 Correct answer: $217$°

(a) $70$°

1. Subtract the two clockwise bearings from the same north line.
   $$125-55$$
2. Therefore $70$°.

• M1 Subtract the two clockwise bearings from the same north line.
• A1 Correct answer: $70$°

(b) $55$°

1. The equal sides make the base angles equal.
   $$\frac{180-70}{2}$$
2. Therefore $55$°.

• M1 The equal sides make the base angles equal.
• A1 Correct answer: $55$°

(a) $135$°

1. Triangle PLB is isosceles and right-angled at L, so B is south-east of P.
   $$180-45$$
2. Therefore $135$°.

• P1 Triangle PLB is isosceles and right-angled at L, so B is south-east of P.
• A1 Correct answer: $135$°

(b) $315$°

1. Add 180° to reverse the bearing.
   $$135+180$$
2. Therefore $315$°.

• M1 Add 180° to reverse the bearing.
• A1 Correct answer: $315$°

(c) $20\sqrt{2}$

1. Use Pythagoras with the two 20 km legs.
   $$\sqrt{20^{2}+20^{2}}$$
2. Therefore $20\sqrt{2}$.

• M1 Use Pythagoras with the two 20 km legs.
• A1 Correct answer: $20\sqrt{2}$

(a) $053.1^\circ$

1. Bearings are measured clockwise from north.
2. Bearings are measured clockwise from north.
3. For the angle θ east of north, tan θ = east/north = 8/6.
4. θ = $\tan^{-1}$(8/6) = $53.130...^\circ$, so the bearing is $053.1^\circ.$

• P1 Bearings are measured clockwise from north.
• C1 Correct conclusion with the complete supporting argument: $053.1^\circ$

(a) $077^\circ$

1. The change of direction at $B$ is $130^\circ - 40^\circ = 90^\circ$, so angle $ABC = 90^\circ$.
2. In the right-angled triangle $ABC$:
   $$\tan(\angle BAC) = \frac{6}{8} \;\Rightarrow\; \angle BAC = 36.87^\circ$$
3. The bearing of $C$ from $A$ is the bearing of $B$ plus this angle.
   $$40 + 36.87 = 76.87$$
4. To the nearest degree: $077^\circ$.

• P1 Showing angle $ABC = 90^\circ$ (from $130 - 40$ and the north lines being parallel).
• P1 Angle $BAC$ from $\tan^{-1}\left(\frac{6}{8}\right)$ (or $AC = 10$ then sine or cosine).
• P1 Adding their angle to $40^\circ$.
• A1 Correct answer: $077^\circ$.

(b) $257^\circ$

1. Reverse the direction: $77 + 180 = 257$.

• B1 Correct answer: $257^\circ$.