October 2021 Paper 1 Q10

OCR ACurrent spec11 marksProof Trigonometry

(a)

Triangle ABC with base AB; the perpendicular CD from C meets AB at D with a right angle; angle ACD is x and angle DCB is y; side AC is b and side BC is a

The diagram shows triangle \(ABC\). The perpendicular from \(C\) to \(AB\) meets \(AB\) at \(D\).
Angle \(ACD = x\), angle \(DCB = y\), length \(BC = a\) and length \(AC = b\).

(i) Explain why the length of \(CD\) can be written as \(a\cos y\). [1]

(ii) Show that the area of the triangle \(ADC\) is given by \(\frac{1}{2}ab\sin x\cos y\). [1]

(iii) Hence, or otherwise, show that \(\sin(x + y) = \sin x\cos y + \cos x\sin y\). [4]

(b) Given that \(\sin(30^\circ + \alpha) = \cos(45^\circ - \alpha)\), show that \(\tan\alpha = 2 + \sqrt{6} - \sqrt{3} - \sqrt{2}\). [5]

Mark scheme (a)

Scheme	Marks	AO
(i) \(\cos y = \frac{CD}{a}\) hence \(CD = a\cos y\)	B1	2.4
	[1]
(ii) area \(= \frac{1}{2}AC.CD\sin x = \frac{1}{2}b(a\cos y)\sin x\) \(= \frac{1}{2}ab\sin x\cos y\) A.G.	B1	2.4
	[1]
(iii) \(CD = b\cos x\)	B1	2.1
Area \(BCD =\) \(\frac{1}{2}BC.CD\sin y = \frac{1}{2}a(b\cos x)\sin y\) \(= \frac{1}{2}ab\cos x\sin y\)	B1	2.1
Area \(ABC =\) \(\frac{1}{2}AC.BC\sin(x + y) = \frac{1}{2}ab\sin(x + y)\)	B1	1.1
\(\frac{1}{2}ab\sin(x + y) = \frac{1}{2}ab\sin x\cos y + \frac{1}{2}ab\cos x\sin y\) \(\sin(x + y) = \sin x\cos y + \cos x\sin y\)	B1	2.1
	[4]

Notes

(a)(i) B1: Justification for \(CD\)
Need to see either \(\cos y = \frac{CD}{a}\) or adj = hyp × \(\cos\theta\) before given answer

(a)(ii) B1: Use area of triangle to show given answer
Could quote general expression for area and then show clear substitution
If not, then sides being used need to be clearly identified through statement or diagram
Could also use right-angled triangle, with base as \(AD\)
Condone not being rearranged to given expression

(a)(iii)

B1: Correct \(CD\) in terms of \(b\) and \(x\)

B1: Correct area of triangle \(BCD\)
B0 B1 if correct area stated with no justification

B1: Correct area of triangle \(ABC\)

B1: Equate area of \(ABC\) to the sum of the areas of the two small triangles and complete proof convincingly
Allow alternative proofs eg using lengths

Mark scheme (b)

Scheme	Marks	AO
\(\sin 30\cos\alpha + \cos 30\sin\alpha =\) \(\cos 45\cos\alpha + \sin 45\sin\alpha\)	B1	1.1
\(\frac{1}{2}\cos\alpha + \frac{1}{2}\sqrt{3}\sin\alpha = \frac{1}{2}\sqrt{2}\cos\alpha + \frac{1}{2}\sqrt{2}\sin\alpha\)	M1	1.1
\(\left(\sqrt{3} - \sqrt{2}\right)\sin\alpha = \left(\sqrt{2} - 1\right)\cos\alpha\) \(\dfrac{\sin\alpha}{\cos\alpha} = \tan\alpha = \dfrac{\sqrt{2} - 1}{\sqrt{3} - \sqrt{2}}\)	M1	3.1a
\(= \dfrac{\left(\sqrt{2} - 1\right)\left(\sqrt{3} + \sqrt{2}\right)}{\left(\sqrt{3} - \sqrt{2}\right)\left(\sqrt{3} + \sqrt{2}\right)} = \dfrac{\sqrt{6} + 2 - \sqrt{2} - \sqrt{3}}{3 - 2}\)	M1	3.1a
\(\tan\alpha = 2 + \sqrt{6} - \sqrt{3} - \sqrt{2}\) A.G.	A1	2.1
	[5]

Notes

B1: Correct use of compound angle formulae
Could be implied if exact values used immediately – allow BOD for RHS
May be seen as two separate expressions, not yet equated

M1: Use exact trig values
In either equation or two expressions
Must see all 4 values, but expansions may not be fully correct

M1: Gather like terms and attempt \(\tan\alpha\)
May still have fractions in the fraction
\(\tan\alpha\) does not yet need to be the subject, but must only appear once for M1

M1: Attempt to rationalise their denominator
Clear intention seen to multiply throughout by the conjugate of their denominator

A1: Obtain given answer
With full detail, including (at least) \(3 - 2\) in denominator