Cosine rule for sides and angles

(a) 7.24 cm

1. Two sides and the angle between them: cosine rule.
2. $$BC^2 = 7^2 + 9^2 - 2 \times 7 \times 9 \cos 52^\circ = 52.43\ldots$$
3. $$BC = 7.2407\ldots$$

• M1 Correct substitution into the cosine rule.
• M1 Square-rooting a correct value of $BC^2$.
• A1 Correct answer: 7.24 cm.

(a) $95.7^\circ$

1. The largest angle is opposite the longest side, 9 cm.
2. Rearranged cosine rule.
   $$\cos\theta = \frac{5^2 + 7^2 - 9^2}{2 \times 5 \times 7} = \frac{-7}{70} = -0.1$$
3. $$\theta = \cos^{-1}(-0.1) = 95.74^\circ$$

• M1 Choosing the angle opposite 9 cm.
• M1 $\cos\theta = \frac{25 + 49 - 81}{70}$.
• A1 Correct answer: $95.7^\circ$.

(b) The cosine is negative, so the angle is more than $90^\circ$.

1. $\cos\theta < 0$ exactly when $90^\circ < \theta < 180^\circ$; equivalently $5^2 + 7^2 < 9^2$.

• C1 A negative cosine (or $5^2 + 7^2 < 9^2$) means the angle is obtuse.

(a) $16.4$ cm

1. $$30/(10\times 8/2)$$
2. $$\sqrt{10^{2}+8^{2}+160\sqrt{7}/4}$$
3. Using area = $\frac{1}{2}ab$ sin θ gives 30 = 40 sin θ, so sin θ = 3/4.
4. Because θ is obtuse, cos θ is negative. From $\sin^{2}$θ + $\cos^{2}$θ = 1, cos θ = $-\sqrt{7}$/4.
5. By the cosine rule, $c^{2}$ = $10^{2}$ + $8^{2}$ $-$ 2 $\times$ 10 $\times$ 8 $\times$ ($-\sqrt{7}$/4) = 164 + $40\sqrt{7}.$
6. c = $\sqrt{}$(164 + $40\sqrt{7}$) = 16.426504... cm, so the third side is 16.4 cm to 3 significant figures.

• P1 Establishing $30/(10\times 8/2)$ or an equivalent valid method.
• P1 Establishing $\sqrt{10^{2}+8^{2}+160\sqrt{7}/4}$ or an equivalent valid method.
• A1 Correct answer: $16.4$ cm

(a) $10.8$ cm

1. Use the cosine rule with the included angle.
   $$8^{2}+5^{2}-2\times 8\times 5\times \cos(111)$$
2. Take the positive square root of the squared side.
   $$\sqrt{8^{2}+5^{2}-2\times 8\times 5\times \cos(111)}$$
3. Therefore $10.8$ cm.

• P1 Use the cosine rule with the included angle.
• P1 Take the positive square root of the squared side.
• A1 Correct answer: $10.8$ cm

(b) BC is opposite the obtuse angle, which is the largest angle of the triangle. The largest side is opposite the largest angle.

1. Therefore BC is opposite the obtuse angle, which is the largest angle of the triangle. The largest side is opposite the largest angle.

• C1 Correct conclusion with supporting reasoning: BC is opposite the obtuse angle, which is the largest angle of the triangle. The largest side is opposite the largest angle.

(a) $85.5$°

1. The largest angle is opposite the longest side; rearrange the cosine rule.
   $$\frac{10^{2}+12^{2}-15^{2}}{2\times 10\times 12}$$
2. Therefore $85.5$°.

• M1 The largest angle is opposite the longest side; rearrange the cosine rule.
• A1 Correct answer: $85.5$°

(b) $59.8$ cm²

1. Use the two sides surrounding the unrounded angle.
   $$0.5\times 10\times 12\times \sin(85.459332671942)$$
2. Therefore $59.8$ cm².

• M1 Use the two sides surrounding the unrounded angle.
• A1 Correct answer: $59.8$ cm²

(a) $86.9$°

1. Use the cosine rule with PR opposite angle Q.
   $$\cos(Q)=(18^{2}+25^{2}-30^{2})/(2\times 18\times 25)$$
2. Therefore $86.9$°.

• M1 Use the cosine rule with PR opposite angle Q.
• A1 Correct answer: $86.9$°

(b) $225$ km²

1. Use one half × PQ × QR × sin Q.
   $$1/2\times 18\times 25\times \sin(86.9)$$
2. Therefore $225$ km².

• M1 Use one half × PQ × QR × sin Q.
• A1 Correct answer: $225$ km²

(a) 10.6 km

1. At $B$, the bearing back to $A$ is $050 + 180 = 230^\circ$.
2. Angle $ABC$ is the difference between the directions $BA$ and $BC$.
   $$230 - 170 = 60$$
3. Cosine rule.
   $$AC^2 = 12^2 + 8^2 - 2 \times 12 \times 8 \cos 60^\circ = 208 - 96 = 112$$
4. $$AC = \sqrt{112} = 10.583\ldots$$

• P1 The back bearing $230^\circ$ (or co-interior angles giving $130^\circ$ between north and $BA$).
• P1 Angle $ABC = 60^\circ$.
• P1 Cosine rule with 12, 8 and their angle.
• A1 Correct answer: 10.6 km.

(b) $091^\circ$

1. Sine rule for angle $BAC$.
   $$\sin(\angle BAC) = \frac{8 \sin 60^\circ}{10.583\ldots} = 0.6547$$
2. $$\angle BAC = 40.89^\circ$$
3. Add to the bearing of $B$.
   $$50 + 40.89 = 90.89$$

• P1 Angle $BAC$ by the sine or cosine rule.
• P1 Adding their angle to $50^\circ$.
• A1 Correct answer: $091^\circ$.

(a) $200$°

1. $$9\times \sin(140)-12\times \sin(60)$$
2. $$9\times \cos(140)-12\times \cos(60)$$
3. Take A as the origin, with east and north as positive coordinates. B has coordinates (12 sin $60^\circ$, 12 cos $60^\circ$) and C has coordinates (9 sin $140^\circ$, 9 cos $140^\circ$).
4. From B to C the east component is $-4.607216...$ km and the north component is $-12.894399...$ km, so C is southwest of B.
5. The angle west of south is $\tan^{-1}$(4.607216.../12.894399...) = $19.661994...^\circ.$
6. The bearing is $180^\circ$ + $19.661994...^\circ$ = $199.661994...^\circ$, giving $200^\circ$ to the nearest degree.

• P1 Establishing $9\times \sin(140)-12\times \sin(60)$ or an equivalent valid method.
• P1 Establishing $9\times \cos(140)-12\times \cos(60)$ or an equivalent valid method.
• A1 Correct answer: $200$°