June 2025 Paper 1 Q16
16 The triangle \(ABC\) is shown in the diagram below.
The angle \(ABC\) is \(\dfrac{\pi}{3}\) radians.
The angle \(BCA\) is \(x\) radians.
(a) Explain why angle \(BAC = \dfrac{2\pi}{3} - x\) [1 mark]
(b)
(i) Use the sine rule to show that\[\frac{BC}{AB} = \frac{\sqrt{3}\cot x + p}{q}\]where \(p\) and \(q\) are integers. [5 marks]
(ii) Hence state the exact value of \(x\) when\[\frac{BC}{AB} = \frac{\sqrt{3} + 1}{2}\] [1 mark]
| Scheme | Marks | AO |
|---|---|---|
| Explains that the sum of the angles in a triangle is \(\boldsymbol{\pi}\) (radians) and verifies \(BAC = \dfrac{2\pi}{3} - x\) Eg \(\pi = \dfrac{2\pi}{3} - x + \dfrac{\pi}{3} + x\) OE | E1 | 2.4 |
| (1) |
Typical solution
The angles in a triangle add up to \(\pi\) radians.
Angle \(BAC = \pi - \dfrac{\pi}{3} - x = \dfrac{2\pi}{3} - x\)
| Scheme | Marks | AO |
|---|---|---|
| (i) Forms an equation using the sine rule using any two of \(\dfrac{AC}{\sin\frac{\pi}{3}} = \dfrac{BC}{\sin A} = \dfrac{AB}{\sin x}\) OE Could use corresponding lowercase letters or letters defined or on the diagram throughout. | M1 | 3.1a |
| Obtains \(\dfrac{BC}{\sin\left(\frac{2\pi}{3} - x\right)} = \dfrac{AB}{\sin x}\) Accept \(\dfrac{BC}{\sin\left(\pi - \left(\frac{\pi}{3} + x\right)\right)} = \dfrac{AB}{\sin x}\) OE | A1 | 1.1b |
| Uses compound angle formula to expand \(\sin\left(\dfrac{2\pi}{3} - x\right)\) or \(\sin\left(\dfrac{\pi}{3} + x\right)\) Condone one sign error | M1 | 3.1a |
| Substitutes correct exact values for \(\sin\left(\dfrac{2\pi}{3}\right)\) and \(\cos\left(\dfrac{2\pi}{3}\right)\) Or for \(\sin\left(\dfrac{\pi}{3}\right)\) and \(\cos\left(\dfrac{\pi}{3}\right)\) into their expanded compound angle | M1 | 1.1a |
| Completes reasoned argument to obtain \(\dfrac{BC}{AB} = \dfrac{\sqrt{3}\cot x + 1}{2}\) Accept \(\dfrac{a}{c} = \dfrac{\sqrt{3}\cot x + 1}{2}\) | R1 | 2.1 |
| (5) | ||
| (ii) Deduces \(\dfrac{\pi}{4}\) from \(\dfrac{\sqrt{3}\cot x + 1}{2}\) | R1 | 2.2a |
| (1) | ||
| (7 marks) |
Typical solution
(b)(i)
\[\frac{BC}{\sin\left(\frac{2\pi}{3} - x\right)} = \frac{AB}{\sin x}\]\[\sin\left(\frac{2\pi}{3} - x\right) = \sin\frac{2\pi}{3}\cos x - \cos\frac{2\pi}{3}\sin x = \frac{\sqrt{3}}{2}\cos x - -\frac{1}{2}\sin x = \frac{\sqrt{3}}{2}\cos x + \frac{1}{2}\sin x\]\[\frac{BC}{AB} = \frac{\frac{\sqrt{3}}{2}\cos x + \frac{1}{2}\sin x}{\sin x} = \frac{\sqrt{3}\cot x + 1}{2}\](b)(ii) \(\dfrac{\pi}{4}\)