Higher January 2022 Paper 2R Q16
16 The diagram shows quadrilateral \(ABCD\)
Diagram NOT accurately drawn
The angle \(BCD\) is acute.
Given that the area of triangle \(BCD = 405\) cm2
work out the size of angle \(ABD\)
Give your answer correct to one decimal place.
(5)
| Scheme | Marks |
|---|---|
\(\dfrac{1}{2} \times 45 \times 36 \times \sin \text{‘}{C}\text{’}\ (= 405)\) or alternative \(\dfrac{2 \times 405}{36}\ (= 22.5)\) or \(\dfrac{2 \times 405}{45}\ (= 18)\) | M1 |
\(\sin \text{‘}{C}\text{’} = \left(\dfrac{405 \times 2}{45 \times 36}\right)(\text{‘}{C}\text{’} = 30)\) oe or \(\sqrt{45^2 - 22.5^2}\left(= \sqrt{1518.75} = 38.97\right)\) or \(\sqrt{36^2 - 18^2}\left(= \sqrt{972} = 31.17\right)\) | M1 |
\((BD =)\sqrt{45^2 + 36^2 - 2 \times 45 \times 36 \times \cos \text{‘}{30}\text{’}}\) \(\left(= \sqrt{3321 - 3240 \times \cos \text{‘}{30}\text{’}}\right)\) \(\left(= \sqrt{515.077\ldots} = 22.695\ldots\right)\) or \(\sqrt{(\text{‘}{38.97}\text{’} - 36)^2 + 22.5^2}\left(= \sqrt{515.077\ldots}\right)\) or \(\sqrt{(\text{‘}{45}\text{’} - 31.17)^2 + 18^2}\left(= \sqrt{515.077\ldots}\right)\) | M1 |
\(\cos \text{‘}{ABD}\text{’} = \left(\dfrac{\text{‘}{22.695\ldots}\text{’}^2 + 19^2 - 28^2}{2 \times \text{‘}{22.695\ldots}\text{’} \times 19}\right)\) leading to ‘\(ABD\)’ = or \((BAD =) \cos^{-1}\left(\dfrac{28^2 + 19^2 - \text{‘}{22.695\ldots}\text{’}^2}{2 \times 28 \times 19}\right)\) (= 53.7…) and \(\sin \text{‘}{ABD}\text{’} = \dfrac{\sin \text{‘}{53.7}\text{’}}{\text{‘}{22.695\ldots}\text{’}} \times 28\) leading to ‘\(ABD\)’ = | M1 |
| 83.9 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: correct substitution into the sine area formula, with their choice of symbol to represent \(C\).
or work out the perpendicular height with \(BC\) or \(CD\) as the base.
M1: correct rearrangement to make \(\sin C\) the subject
or use Pythagoras with their found perpendicular height.
M1: (dep on 1st M1, ft 30) correct expression for \(BD\) ft their \(C\) (must be less than 90°).
or use Pythagoras to find an expression for \(BD\).
M1: for a complete method to find angle \(ABD\)
A1: accept 83.85 – 83.9