A-LevelMathematicsTrigonometry9709/12 · May/June 2016 · Q55 Marks
In the diagram, triangle ABC is right-angled at C and M is the mid-point of BC. It is given that angle ABC = 1/3 π radians and angle BAM = θ radians. Denoting the lengths of BM and MC by x, [Figure 5.1]
- (i)find AM in terms of x,[3]
- (ii)show that θ = 1/6 π – tan⁻¹(1/(2√3))[2]
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Mark scheme
Marking points from the official mark scheme (May/June 2016). The mark or mark type is shown in brackets.
- [B1] Either trig ratio
- tan(π/6) = AC/(2x) or cos(π/6) = (2x)/AB or sin(π/6) = AC/AB
- → AC = 2√3x or AB = 4x
- [M1A1] AM = √13x², √13x, 3.61x Complete method.
- [M1] tan(MAC) = x / Their AC "Their AC" must be f(x), (MAC)≠0.
- [A1] θ = 1/6 π – tan⁻¹(1/(2√3)) AG Justifies π/6 and links MAC & θ
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