Higher January 2021 Paper 1 Q10
10 Here is isosceles triangle \(ABC\).
Diagram NOT accurately drawn
\(D\) is the midpoint of \(AC\) and \(DB\) = 16 cm.
Angle \(DAB = 65^\circ\)
Work out the perimeter of triangle \(ABC\).
Give your answer correct to one decimal place.
(4)
| Scheme | Marks |
|---|---|
e.g. \(\sin 65 = \dfrac{16}{AB}\) or \(\cos 25 = \dfrac{16}{AB}\) or \(\dfrac{AB}{\sin 90} = \dfrac{16}{\sin 65}\) or \(\tan 65 = \dfrac{16}{AD}\) or \(\tan 25 = \dfrac{AD}{16}\) or \(\dfrac{AD}{\sin 25} = \dfrac{16}{\sin 65}\) | M1 |
e.g. \((AB =)\;\dfrac{16}{\sin 65}\) (= 17.654…) or \((AB =)\;\dfrac{16}{\cos 25}\) (= 17.654…) or \((AB =)\;\dfrac{16\sin 90}{\sin 65}\) (= 17.654…) and \((AD =)\;\dfrac{16}{\tan 65}\) (= 7.460…) or \((AD) = 16 \times \tan 25\) (= 7.460…) or \((AD =)\;\dfrac{16\sin 25}{\sin 65}\) (= 7.460…) | M1 |
| (“17.654…” × 2) + (“7.460…” × 2) oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 50.2 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for a correct trig ratio for \(AB\) or \(AD\)
accept 180 – 90 – 65 for 25
M1: for finding \(AB\) and \(AD\)
Allow use of Pythagoras
\((AD =)\sqrt{\text{“}17.654...\text{”}^2 - 16^2}\) (= 7.460…)
or
\((AB =)\sqrt{\text{“}7.460...\text{”}^2 + 16^2}\) (= 17.654…)
M1: for a complete method to find the perimeter
A1: accept 49.6 – 50.6