Higher June 2023 Paper 4 Q20
20 The diagram shows triangle ABC.
Not to scale
AB = 10.6 cm, BC = 8.2 cm and AC = 12.5 cm.
(a) Show that angle BAC = 40.5°, correct to 1 decimal place. [3]
(b) Work out the area of triangle ABC. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [cos BAC = ] \(\frac{10.6^2 + 12.5^2 - 8.2^2}{2 \times 10.6 \times 12.5}\) oe or better | M2 | M1 for \(8.2^2 = 10.6^2 + 12.5^2 - 2 \times 10.6 \times 12.5 \times \cos[.]\) oe and M1dep for 0.7598… or 265 cos[.] = 201.37 | Do not reward work from use of 40.5°. M2 implied by \(\frac{201.37}{265}\) oe and accept 67.24 for \(8.2^2\) etc Dep. on previous M1 |
| 40.54… | A1 | Dep. on at least M1 scored If 0 scored award SC1 for 40.54… | For M2 and M1 accept alternative methods which must be correct and complete. |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 43.02 to 43.07 or 43[.0] or 43.1 | 2 | M1 for \(\frac{1}{2}\) × 10.6 × 12.5 × sin 40.5… | For M1 accept alternative methods which must be correct and complete e.g using a different angle |