Higher June 2022 Paper 2 Q24
24.
Diagram not drawn to scale
\(A\hat{B}C\) is acute.
The area of triangle ABC is 45·0 cm2.
Show that \(A\hat{B}C = 50^\circ\) correct to the nearest integer and hence calculate \(A\hat{D}C\). [9]
| Scheme | Marks | Comments |
|---|---|---|
| Degree symbol may be omitted throughout | ||
| \(\sin(\ldots) = \dfrac{45.0 \times 2}{12.5 \times 9.4}\) oe | M2 | M1 for \(\dfrac12 \times 12.5 \times 9.4 \times \sin ABC = 45.0\) oe |
| \(A\hat{B}C = 49.9(922\ldots)^\circ\ (= 50^\circ)\) | A1 | If M0 awarded, then: SC2 for \(\dfrac12 \times 12.5 \times 9.4 \times \sin 50 = 45(.005\ldots)\) or SC1 for \(\dfrac12 \times 12.5 \times 9.4 \times \sin 50\) |
| Alternative method for the first 3 marks | ||
| (Perp height from C to AB =) \(45 \times 2 \div 12.5\) oe | M1 | Allow for appropriate sight of 7.2 (cm) |
| \(\sin A\hat{B}C = \dfrac{7.2}{9.4}\) | M1 | |
| \(A\hat{B}C = 49.9(922\ldots)^\circ\ (= 50^\circ)\) | A1 | |
| \(AC = \sqrt{12.5^2 + 9.4^2 - 2(12.5)(9.4)\cos 50}\) | M2 | Accept answers in range \(\sqrt{93.2}\) to \(\sqrt{93.6}\) M1 for \(AC^2 = 12.5^2 + 9.4^2 - 2(12.5)(9.4)\cos 50\) implied by 93.2 to 95.6 |
| AC = 9.67 to 9.7 (cm) si | A1 | Allow 10 cm from correct working Allow A1 for \(\sqrt{93.2}\) to \(\sqrt{93.6}\) |
| \(\sin A\hat{D}C = \dfrac{9.67\ldots \times \sin 85}{10.2}\ (= 0.9446\ldots)\) oe | M2 | FT ‘their 9.67...’ M1 for \(\dfrac{9.67\ldots}{\sin ADC} = \dfrac{10.2}{\sin 85}\) oe |
| Accept answers in range 70.8° to 71.33° | A1 | CAO |
| (9) | ||
Corrected by Westie's Workshop
- The board's mark scheme printed “sin(...) = 0.7 (45.0 × 2)/(12.5 × 9.4)”. Changed to “sin(...) = (45.0 × 2)/(12.5 × 9.4)” because the stray 0.7 is a printing leftover and would make the line wrong.
Original: see the board's mark scheme PDF (link above).