Higher June 2023 Paper 2 Q22
22 The diagram shows a triangle \(ABC\) and a flagpole \(BF\)
Diagram NOT accurately drawn
\(A\), \(B\) and \(C\) are points on horizontal ground.
\(BF\) is vertical.
\(AB = 9\) m \(\qquad BC = 11\) m \(\qquad AC = 16\) m \(\qquad BF = 10\) m
\(D\) is the point on \(AC\) such that angle \(BDC = 90^\circ\)
Work out the size of the angle of elevation of the point \(F\) from the point \(D\)
Give your answer correct to one decimal place.
(5)
| Scheme | Marks |
|---|---|
\(9^2 = 11^2 + 16^2 - 2 \times 11 \times 16 \times \cos BCA\) oe or \(11^2 = 9^2 + 16^2 - 2 \times 9 \times 16 \times \cos BAC\) or \(16^2 = 9^2 + 11^2 - 2 \times 9 \times 11 \times \cos ABC\) or (area of \(\triangle ABC\) =) \(\sqrt{18 \times 2 \times 7 \times 9}\) (= 47.6235...) oe | M1 |
\((\cos BCA =)\left(\dfrac{11^2 + 16^2 - 9^2}{2 \times 11 \times 16}\right)\) (\(BCA\) = 32.763...) or \((\cos BAC =)\left(\dfrac{9^2 + 16^2 - 11^2}{2 \times 9 \times 16}\right)\) (\(BAC\) = 41.409...) or \((\cos ABC =)\left(\dfrac{9^2 + 11^2 - 16^2}{2 \times 9 \times 11}\right)\) (\(ABC\) = 105.826...) or \(\dfrac{1}{2} \times 16 \times BD = \text{``}{47.6235...}\text{''}\) | M1 |
(\(BD\) =) 11 sin”32.763...”(= 5.95...) oe eg 11 sin(180 – “41.4…” – 105.8…”) (= 5.95…) or 9 sin”41.4….” (= 5.95…) oe or \(\dfrac{\text{``}{47.6235...}\text{''} \times 2}{16}\) (=5.95...) oe or \(\sqrt{11^2 - \text{``}{9.25}\text{''}^2}\) or \(\sqrt{9^2 - \text{``}{6.75}\text{''}^2}\) \(11\sin\left(\sin^{-1}\left(\dfrac{9\sin\text{``}{105.826...}\text{''}}{16}\right)\right)\) (= 5.95…) oe | M1 |
| \(\tan FDB = \dfrac{10}{\text{``}{5.95...}\text{''}}\) oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 59.2 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: For a start to the correct method to find angle \(BCA\) or angle \(BAC\) or angle \(ABC\) or a fully correct method to find the area of the triangle
M1: For a correct rearrangement for \(\cos BCA\) or \(\cos BAC\) or \(\cos ABC\) or a correct equation to find \(BD\)
(accept angles to the nearest whole number rounded or truncated as long as not from incorrect working)
M1: For a correct calculation that will lead to the value of \(BD\)
“47.6235…” may also come from
0.5 × 9 × 11 × sin”105.8…” or
0.5 × 9 × 16 × sin”41.4…” or
0.5 × 16 × 11 × sin”32.7…”
[Students may find an angle by sine rule after already finding an angle and use this]
M1: For a correct expression for the required angle (in form \(\tan x\) =… or \(\cos x\) =… or \(\sin x\) =… ) oe
A1: awrt 59.2
22 ALT
Angle \(DBC\) = 57.237… Angle \(ABD\) = 48.591… \(AD\) = 6.75 m \(\quad CD\) = 9.25 m
| Scheme | Marks |
|---|---|
| \(BD^2 = 11^2 - (16 - y)^2\) and \(BD^2 = 9^2 - y^2\) oe | M1 |
| \(11^2 - (16 - y)^2 = 9^2 - y^2\) (\(y\) = 6.75 or \(x\) = 9.25) | M1 |
| \(BD = \sqrt{9^2 - (16 - \text{``}{9.25}\text{''})^2}\) or \(\sqrt{11^2 - \text{``}{9.25}\text{''}^2}\) (= 5.95) | M1 |
| \(\tan FDB = \dfrac{10}{\text{``}{5.95...}\text{''}}\) oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 59.2 | A1 |
| (5 marks) |
Notes
M1: For 2 different expressions in the same single variable for \(BD\) or \(BD^2\)
M1: Equating the 2 expressions
M1: A correct calculation to find BD
(“9.25” or “6.75” must come from a correct method)
M1: For a correct expression for the required angle (in form \(\tan x\) =… or \(\cos x\) =… of \(\sin x\) =… ) oe
A1: awrt 59.2