Sine rule and the ambiguous case

(a) 11.3 cm

1. $BC$ is opposite angle $A$ and $AC$ is opposite angle $B$.
2. Sine rule.
   $$\frac{AC}{\sin 65^\circ} = \frac{8}{\sin 40^\circ}$$
3. $$AC = \frac{8 \sin 65^\circ}{\sin 40^\circ} = 11.279\ldots$$

• M1 A correct sine-rule statement pairing each side with its opposite angle.
• A1 Correct answer: 11.3 cm.

(a) $14.0$ cm

1. $$8\times \sin(79)/\sin(34)$$
2. BC is opposite angle A and AC is opposite angle B.
3. By the sine rule, AC/sin $79^\circ$ = 8/sin $34^\circ.$
4. AC = 8 sin $79^\circ$/sin $34^\circ$ = 14.043485... cm, which is 14.0 cm to 3 significant figures.

• P1 Establishing $8\times \sin(79)/\sin(34)$ or an equivalent valid method.
• A1 Correct answer: $14.0$ cm

(a) $6\sqrt{2}$ cm

1. $YZ$ is opposite $X$ and $XZ$ is opposite $Y$.
2. $$XZ = \frac{6 \sin 45^\circ}{\sin 30^\circ} = \frac{6 \times \frac{\sqrt{2}}{2}}{\frac{1}{2}} = 6\sqrt{2}$$

• M1 A correct sine-rule statement.
• M1 Substituting the exact values $\frac{\sqrt{2}}{2}$ and $\frac{1}{2}$.
• A1 Correct answer: $6\sqrt{2}$ cm.

(a) $15.2$ cm

1. Pair each side with its opposite angle in the sine rule.
   $$9\times \sin(71)/\sin(34)$$
2. Therefore $15.2$ cm.

• M1 Pair each side with its opposite angle in the sine rule.
• A1 Correct answer: $15.2$ cm

(b) $75$°

1. The three interior angles sum to 180 degrees.
   $$180-34-71$$
2. Therefore $75$°.

• M1 The three interior angles sum to 180 degrees.
• A1 Correct answer: $75$°

(a) 48.6 degrees and 131.4 degrees.

1. Apply the sine rule to obtain sin B.
   $$12\times \sin(30)/8$$
2. Use the acute inverse-sine angle and its supplement.
   $$180-48.5903778907$$
3. Therefore 48.6 degrees and 131.4 degrees.

• P1 Apply the sine rule to obtain sin B.
• P1 Use the acute inverse-sine angle and its supplement.
• A1 Correct answer: 48.6 degrees and 131.4 degrees.

(b) $47.1$ cm²

1. Find each included angle C and compare the resulting triangle areas.
   $$0.5\times 8\times 12\times \sin(101.409622109)$$
2. Therefore $47.1$ cm².

• P1 Find each included angle C and compare the resulting triangle areas.
• A1 Correct answer: $47.1$ cm²

(a) No. The sine rule gives sin B = 18 sin 35 / 10, approximately 1.032. A real angle cannot have sine greater than 1.

1. Pair the known side and angle and calculate the implied other sine.
   $$18\times \sin(35)/10$$
2. No. The sine rule gives sin B = 18 sin 35 / 10, approximately 1.032. A real angle cannot have sine greater than 1.

• M1 Pair the known side and angle and calculate the implied other sine.
• C1 Correct conclusion with supporting reasoning: No. The sine rule gives sin B = 18 sin 35 / 10, approximately 1.032. A real angle cannot have sine greater than 1.

(b) No. Neither an angle nor its supplement can have sine greater than 1. There is no real inverse-sine value here.

1. No. Neither an angle nor its supplement can have sine greater than 1. There is no real inverse-sine value here.

• C1 Correct conclusion with supporting reasoning: No. Neither an angle nor its supplement can have sine greater than 1. There is no real inverse-sine value here.

(a) $84.3^\circ$

1. Sine rule for angle $B$ ($AC$ is opposite $B$).
   $$\sin B = \frac{9 \sin 40^\circ}{7} = 0.8264\ldots$$
2. The acute solution.
   $$B = 55.74^\circ$$
3. Angles in a triangle.
   $$C = 180 - 40 - 55.74 = 84.26^\circ$$

• P1 $\sin B = \frac{9 \sin 40^\circ}{7}$.
• P1 $B = 55.7^\circ$.
• A1 Correct answer: $84.3^\circ$.

(b) $15.7^\circ$

1. The obtuse solution for $B$ has the same sine.
   $$B = 180 - 55.74 = 124.26^\circ$$
2. It fits: $40 + 124.26 = 164.26 < 180$.
3. $$C = 180 - 40 - 124.26 = 15.74^\circ$$

• P1 The obtuse value $B = 124.3^\circ$, checked against the angle sum.
• P1 Using the angle sum with the obtuse $B$.
• A1 Correct answer: $15.7^\circ$.

(a) $53.8$ cm²

1. $$12\times \sin(35)/9$$
2. The two possible angles C are $49.886408...^\circ$ and $130.113591...^\circ.$ Hence angle A is $95.113591...^\circ$ or $14.886408...^\circ.$
3. Use the sine rule: sin C/12 = sin $35^\circ$/9, so sin C = 12 sin $35^\circ$/9.
4. The two possible angles C are $49.886408...^\circ$ and $130.113591...^\circ.$ Hence angle A is $95.113591...^\circ$ or $14.886408...^\circ.$
5. Area = $\frac{1}{2}$ $\times$ AB $\times$ AC $\times$ sin A = 54 sin A. The areas are approximately 53.7851 cm² and 13.8728 cm².
6. The larger area is 53.8 cm² to 3 significant figures.

• P1 Establishing $12\times \sin(35)/9$ or an equivalent valid method.
• P1 The two possible angles C are $49.886408...^\circ$ and $130.113591...^\circ.$ Hence angle A is $95.113591...^\circ$ or $14.886408...^\circ.$
• A1 Correct answer: $53.8$ cm²