Area of any triangle and mixed trigonometry

(a) $32.2\text{ cm}^2$

1. Area $= \frac{1}{2}a b\sin C$ with the included angle.
   $$\tfrac{1}{2} \times 8 \times 11 \times \sin 47^\circ = 32.18\ldots$$

• M1 $\frac{1}{2} \times 8 \times 11 \times \sin 47^\circ$.
• A1 $32.2\text{ cm}^2$.

(a) $27.6$ cm²

1. $$7\times 10\times \sin(128)/2$$
2. For two sides and the included angle, area = $\frac{1}{2}ab$ sin C.
3. Area = $\frac{1}{2}$ $\times$ 7 $\times$ 10 $\times$ sin $128^\circ$ = 27.580376... cm².
4. To 1 decimal place, the area is 27.6 cm².

• P1 Establishing $7\times 10\times \sin(128)/2$ or an equivalent valid method.
• A1 Correct answer: $27.6$ cm²

(a) $12\sqrt{3}$

1. Use $\frac{1}{2}ab\sin C$ with $\sin 60^\circ = \frac{\sqrt{3}}{2}$.
   $$\frac{1}{2}\times 6\times 8\times \frac{\sqrt{3}}{2}$$
2. Therefore $12\sqrt{3}$.

• M1 Use one half ab sine C and the exact sine of 60 degrees.
• A1 Correct answer: $12\sqrt{3}$

(b) The areas are equal because sin 120 degrees = sin 60 degrees.

1. Supplementary angles have equal sines.
2. The areas are equal because sin 120 degrees = sin 60 degrees.

• M1 Supplementary angles have equal sines.
• C1 Correct conclusion with supporting reasoning: The areas are equal because sin 120 degrees = sin 60 degrees.

(a) $143.1$°

1. Rearrange the area formula to find the sine.
   $$2\times 18.9/(7\times 9)$$
2. Select the supplementary obtuse angle.
   $$180-36.869897645844$$
3. Therefore $143.1$°.

• P1 Rearrange the area formula to find the sine.
• P1 Select the supplementary obtuse angle.
• A1 Correct answer: $143.1$°

(b) $5.4$ cm

1. Use the area as one half base times perpendicular height.
   $$2\times 18.9/7$$
2. Therefore $5.4$ cm.

• P1 Use the area as one half base times perpendicular height.
• A1 Correct answer: $5.4$ cm

(a) $\frac{49}{4}\pi -49/2$

1. Find the quarter-circle sector area.
   $$\pi \times 7^{2}/4$$
2. Subtract the right triangle formed by the two radii.
   $$\pi \times 7^{2}/4-7^{2}/2$$
3. Therefore $\frac{49}{4}\pi -49/2$.

• P1 Find the quarter-circle sector area.
• P1 Subtract the right triangle formed by the two radii.
• A1 Correct answer: $\frac{49}{4}\pi -49/2$

(b) $7\sqrt{2}$

1. Use Pythagoras on the two perpendicular radii.
   $$\sqrt{7^{2}+7^{2}}$$
2. Therefore $7\sqrt{2}$.

• P1 Use Pythagoras on the two perpendicular radii.
• A1 Correct answer: $7\sqrt{2}$

(a) $27.8\text{ m}^2$

1. Find an angle first, for example the one between 7 and 8 (opposite 10).
   $$\cos C = \frac{49 + 64 - 100}{2 \times 7 \times 8} = \frac{13}{112}$$
2. $$C = 83.33^\circ$$
3. Area with the included angle.
   $$\tfrac{1}{2} \times 7 \times 8 \times \sin 83.33^\circ = 27.81\ldots$$

• P1 An angle from the cosine rule, with the correct opposite side.
• P1 Using $\frac{1}{2}a b\sin C$ with the two sides that enclose their angle.
• A1 $27.8\text{ m}^2$.

(a) $60, 120$

1. Use one half × 12 × 9 × sin A for the area.
   $$1/2\times 12\times 9\times \sin(A)=27\sqrt{3}$$
2. Therefore $60, 120$.

• M1 Use one half × 12 × 9 × sin A for the area.
• A1 Correct answer: $60, 120$

(b) $3\sqrt{37}$

1. Use the cosine rule with cos 120° = −1/2.
   $$12^{2}+9^{2}-2\times 12\times 9\times \cos(120)$$
2. Therefore $3\sqrt{37}$.

• M1 Use the cosine rule with cos 120° = −1/2.
• A1 Correct answer: $3\sqrt{37}$

(a) $\frac{100\pi}{3} - 25\sqrt{3}$

1. $$\frac{120}{360}\pi \times 10^{2}$$
2. $$10\times 10\times \sin(120)/2$$
3. The minor segment is the $120^\circ$ sector with triangle OAB removed.
4. Sector area = (120/360) $\times$ $\pi$ $\times$ $10^{2}$ = $100\pi$/3 cm².
5. Triangle area = $\frac{1}{2}$ $\times$ 10 $\times$ 10 $\times$ sin $120^\circ$ = 50 $\times$ $\sqrt{3}$/2 = $25\sqrt{3}$ cm².
6. Subtract to obtain $100\pi$/3 $-$ $25\sqrt{3}$ cm².

• P1 Establishing $\frac{120}{360}\pi \times 10^{2}$ or an equivalent valid method.
• P1 Establishing $10\times 10\times \sin(120)/2$ or an equivalent valid method.
• A1 Correct answer: $\frac{100\pi}{3} - 25\sqrt{3}$