4MA0 Higher June 2017 Paper 4R Q13
13 The diagram shows triangle \(ABC\).
Diagram NOT accurately drawn
\(AB = 9\) cm \(\quad BC = 15\) cm
\(D\) is the point on \(AC\) such that \(AD = 5\) cm.
Angle \(BAC = 90^\circ\)
Calculate the size of angle \(x\).
Give your answer to the nearest degree.
(4)
| Scheme | Marks |
|---|---|
(Angle \(ABD =\)) \(\tan^{-1}\left(\dfrac{5}{9}\right)\) or (Angle \(ADB =\)) \(\tan^{-1}\left(\dfrac{9}{5}\right)\) or (Angle \(ABC =\)) \(\cos^{-1}\left(\dfrac{9}{15}\right)\) or (Angle \(ACB =\)) \(\sin^{-1}\left(\dfrac{9}{15}\right)\) (\(BD =\)) \(\sqrt{9^2 + 5^2}\) or (\(AC =\)) \(\sqrt{15^2 - 9^2}\) (\(DC =\)) \(\sqrt{15^2 - 9^2} - 5\) | M1 |
| A1 | |
Eg (\(x =\)) \(\cos^{-1}\left(\dfrac{9}{15}\right) - \tan^{-1}\left(\dfrac{5}{9}\right)\) or (\(x =\)) 180 − 90 − “29.(0546...)” − “36.(8698...)” or \(\cos x = \dfrac{15^2 + \text{“}10.(2956...)\text{”}^2 - 7^2}{2 \times 15 \times \text{“}10.(2956...)\text{”}}\) or \(\cos x\) = 0.913(009) or \(\sin x = \dfrac{7\sin\text{“}36.(8698...)\text{”}}{\sqrt{106}}\) or \(\sin x = \dfrac{7\sin\text{“}119.(054...)\text{”}}{15}\) or \(\sin x\) = 0.407(940...) | M1 |
| 24 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: For correct method to find angle \(ABD\) or \(ADB\) or \(ABC\) or \(ACB\) or for correct method to find side \(BD\), \(AC\) or \(DC\).
A1: For angle \(ABD\) = 29.(0546...) or
for angle \(ADB\) = 60.(9453...) or
for angle \(ABC\) = 53.(1301...) or
for angle \(ACB\) = 36.(8698...) or
For \(BD = \sqrt{106}\) or 10.(2956...) or
for \(AC\) = 12 or
for \(DC\) = 7
Accept rounded or truncated to at least 2SF
M1: For a complete method to find \(x\) or \(\sin x\) or \(\cos x\)
Accept \(0.912 \leqslant \cos x \leqslant 0.9152\)
Accept \(0.407 \leqslant \sin x \leqslant 0.413\)
A1: Awrt 24