Higher November 2021 Paper 2 Q10
10 \(ABC\) is an isosceles triangle with \(BA = BC\).
Diagram NOT accurately drawn
\(N\) is the point on \(AC\) such that \(AN = 9.3\) cm and \(BN\) is perpendicular to \(AC\).
Work out the perimeter of triangle \(ABC\).
Give your answer correct to 3 significant figures.
(4)
| Scheme | Marks |
|---|---|
eg \(\cos 38 = \dfrac{9.3}{(AB)}\) oe or \(\sin \text{‘}{52}\text{’} = \dfrac{9.3}{(AB)}\) oe or \(\dfrac{(BC)}{\sin 38} = \dfrac{2 \times 9.3}{\sin \text{‘}{104}\text{’}}\) oe or \(\dfrac{\sin \text{‘}{52}\text{’}}{9.3} = \dfrac{\sin 90}{(BC)}\) oe | M1 |
eg \((AB =)\ \dfrac{9.3}{\cos 38}\) (= 11.80….) or \((AB =)\ \dfrac{9.3}{\sin \text{‘}{52}\text{’}}\) (= 11.80….) or \((BC =)\ \dfrac{2 \times 9.3 \times \sin 38}{\sin \text{‘}{104}\text{’}}\) (= 11.80…) oe | M1 |
| ‘11.8’ + ‘11.8’ + 9.3 + 9.3 or ‘11.8’ × 2 + 9.3 × 2 oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 42.2 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: or
\(BN = \dfrac{9.3 \sin 38}{\sin \text{‘}{52}\text{’}}\) or \(9.3 \tan 38\ (= 7.2659\ldots)\)
and
\((AB^2) = 9.3^2 + \text{‘}{7.2659\ldots}\text{’}^2\)
M1: or
\((AB =)\ \sqrt{9.3^2 + \text{‘}{7.2659\ldots}\text{’}^2}\ (= 11.80\ldots)\)
A1: awrt 42.2